Download MAT 1015 Calculus I 2010/2011 John F. Rayman

MAT 1015
Calculus I
2010/2011
John F. Rayman
Department of Mathematics
University of Surrey
Contents
1 Complex numbers
6
1.1
The square root of −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.2
Algebra of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.3
The Argand diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.4
Polar form of complex numbers . . . . . . . . . . . . . . . . . . . . . . . .
9
1.5
De Moivre’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.6
Sets of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2 Functions
15
2.1
Domain and range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.2
Types of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.3
Some important functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.4
Composition of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.5
Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.6
Partial fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.7
Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.8
The exponential and logarithmic functions . . . . . . . . . . . . . . . . . . 26
2.9
Hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.10 Functions of a complex variable . . . . . . . . . . . . . . . . . . . . . . . . 30
3 Limits, graphs and equations
35
3.1
Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.2
Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.3
Standard transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.4
Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.5
Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
4 Differentiation
49
1
4.1
Derivatives of Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . 50
4.2
Differentiating Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . 51
4.3
Parametric Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.4
Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
4.5
Logarithmic Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.6
Leibniz’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.7
Derivatives of functions in polar coordinates . . . . . . . . . . . . . . . . . 54
4.8
Applications of differentials . . . . . . . . . . . . . . . . . . . . . . . . . . 55
5 Series
60
5.1
Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
5.2
Taylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
6 Integration
69
6.1
The anti-derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.2
The definite integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
6.3
The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . 70
6.4
Properties of Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 71
6.5
Integration techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
6.6
Reduction Formulae
6.7
Expressions that cannot be integrated to closed form functions . . . . . . . 79
6.8
Integration and polar curves . . . . . . . . . . . . . . . . . . . . . . . . . . 80
6.9
Applications of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
6.10 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
7 First order ordinary differential equations
87
7.1
Types of first order o.d.e.s . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
7.2
Variables separable equations . . . . . . . . . . . . . . . . . . . . . . . . . 87
7.3
Linear equations - the integrating factor method . . . . . . . . . . . . . . . 88
7.4
Homogeneous equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
7.5
Bernoulli equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
7.6
Equations that can be transformed to one of the types above . . . . . . . . 92
7.7
Second order equations which can be solved as first order equations . . . . 94
7.8
First order o.d.e.s with degree greater than 1 . . . . . . . . . . . . . . . . . 96
7.9
Using Taylor series to obtain a series solution . . . . . . . . . . . . . . . . 98
7.10 Modeling with first order odes . . . . . . . . . . . . . . . . . . . . . . . . . 99
2
8 Introduction to Fourier Series
105
8.1
Definition of a Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . 105
8.2
Integrating odd and even functions . . . . . . . . . . . . . . . . . . . . . . 106
8.3
Complex Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
8.4
Fourier series for functions given on one interval only . . . . . . . . . . . . 112
8.5
Convergence of the Fourier series . . . . . . . . . . . . . . . . . . . . . . . 114
3
Calculus I
Autumn Semester 2010-2011
The calculus module runs for both Autumn and Spring semesters. These are the notes for
Autumn semester. The assessment for the Autumn semester is based on two class tests,
the first counting for 15% and the second 25% and an examination which counts for the
remaining 60%. The Autumn semseter counts for 50% of the total module marks.
There will be three courseworks which will be marked but not assessed.
Contacting the lecturer
My office is 16AA04, my internal telephone extension 2637. If I am not in my office please
feel free to e-mail me on [email protected] to make an appointment. My website is
http://personal.maths.surrey.ac.uk/st/J.Rayman.
Notes
These notes are issued with numerous gaps which will be completed during lectures. Supplementary material will also be distributed from time to time.
Lecture attendance is therefore essential to gain a full understanding of the
material.
Exercises
There are exercises at the end of each chapter and solutions will be distributed progressively. Slightly more challenging exercises are marked with an asterisk.
Background material
While these notes contain all the material you will need to cover during the Autumn
semester there are numerous excellent calculus textbooks in the library. Although they
4
contain far more material than will be covered during the Autumn semester they provide
interesting and useful background and you are encouraged to look at the early chapters
of some of them.
In the University Library, the section coded 51 contains books on ‘General Mathematics’.
Books specifically on Calculus can be found in the section coded 517.
• Guide2 Mathematical methods, by John Gilbert and Camilla Jordan, Palgrave
• Calculus - a complete course, by Robert A. Adams, Addison Wesley
• Calculus - late transcendentals, by Howard Anton et.al, John Wiley and Sons
• Calculus, by Ron Larsen and Bruce Edwards, Brooks/Cole.
The latter three books are very comprehensive and cover all the calculus you are likely to
need for your degree.
5
Chapter 1
Complex numbers
1.1
The square root of −1
A quadratic equation x2 + ax + b = 0 may or may not have solutions in R. Consider
the equation x2 + 4x + 5 = 0. If its roots are α and β then by the theory of quadratic
equations, α + β = −4, αβ = 5.
√
√
Solving the equation gives x√ = −2 ± −1. We√know that −1 does not exist in
R. However,
if α √
= −2 + −1√and β √
= −2 −√ −1, then α + β = −4 and αβ =
√
(−2 + −1)(−2 − −1) = 4 + 2 −1 − 2 −1 − ( −1)2 = 4 − (−1) = 5.
We define i such that i2 = −1. An expression of the form z = x + yi, where x, y ∈ R, is
called a complex number. If x = 0 then z = yi is called a purely imaginary number.
The set of all complex numbers is denoted by C. Clearly R ⊂ C.
x is called the real part of z, Re(z). y is called the imaginary part of z, Im(z).
1.2
Algebra of complex numbers
Two complex numbers are defined to be equal if their real parts are equal and their
imaginary parts are equal,
Example 1
x + yi = 3 − 5i
We do arithmetic in C by treating a complex number as a linear function of i, where
i2 = −1. Thus
6
Example 2
(a + bi) + (c + di) =
(a + bi)(c + di) =
It can be shown that all the usual rules of arithmetic apply to complex numbers, e.g. if
z1 , z2 , z3 ∈ C then
z1 + z2 = z2 + z1
addition is commutative
z1 (z2 + z3 ) = z1 z2 + z1 z3 multiplication is distributive over addition
(z1 z2 )z3 = z1 (z2 z3 )
multiplication is associative
1.2.1
Complex conjugates
z = x − yi is called the conjugate of z = x + yi. It is sometimes denoted by z ? . Note
that zz = x2 + y 2 , which is a real number.
Division of complex numbers is carried out by making the denominator real:
w
wz
=
.
z
zz
Example 3
z=
a + bi
c + di
7
1.2.2
Some useful properties of complex numbers
It is straightforward to prove the following:
Re(z + w)
Im(z + w)
|R(z)|
|Im(z)|
z + z̄
z − z̄
|z| = |z̄|
z±w
z̄¯
zw
(z)
w
|zw|
z
w
|z + w|
1.3
=
=
≤
≤
=
=
=
=
=
=
=
=
=
≤
Re(z) + Re(w)
Im(z) + Im(w)
|z|
|z|
2Re(z)
2Im(z)
| − z| = |−z|
z̄ ± w̄
z
z̄ w̄
z̄
w̄
|z||w|
|z|
|w|
|z| + |w| (the triangle inequality)
The Argand diagram
The complex number z = x + yi can be represented by the point (x, y) in an Argand
diagram or complex plane. Then addition and subtraction correspond to the same
operations with vectors. The graph shows that z, −z, z, −z lie on a circle.
imaginary
−
−z=−x+yi
z=x+yi
real
−
z=x−yi
−z=−x−yi
8
Example 4
Find
√
i
This method can be used to find the square roots of any complex number.
1.4
Polar form of complex numbers
A point in two-dimensions can be specified by its cartesian coordinates (x, y) or its polar
coordinates (r, θ). If z = x + yi, then r and θ determine the polar form of z.
The modulus of z = x + yi is |z| = r =
√
x2 + y 2 . Thus zz = |z|2 .
The argument of z = x + yi is arg(z) = θ where tan θ = xy . In order to obtain the
correct value of the argument it is always a good idea to sketch the Argand diagram.
The principal value of the argument is in the interval (−π, π], so we take 0 < θ ≤ π if
y > 0 and −π < θ < 0 if y < 0. Note that the argument is a many-valued function so
that we could write
arg(1 + i) =
π
+ 2nπ,
4
n = 0, ±1, ±2 . . .
In polar form, the complex number with modulus r and argument θ is r(cos θ + i sin θ).
9
imaginary
r sin θ
θ =π
θ = −π
imaginary
r
θ
θ= 0
r cos θ
real
π+ tan −1y /x
tan −1y /x
real
−π+ tan −1y
/x
tan −1y/x
Example 5
Express z = 2 + 2i and z = 4 − 4i in polar form.
1.5
De Moivre’s Theorem
Let z1 = r1 (cos θ1 + i sin θ1 ), z2 = r2 (cos θ2 + i sin θ2 ). Then, using the trigonometric addition formulae, we have z1 z2 = r1 r2 (cos(θ1 + θ2 ) + i sin(θ1 + θ2 )). Thus z1 z2 has modulus
r1 r2 (multiply the moduli) and argument (θ1 + θ2 ). (add the arguments)
z1
r1
z1
Dividing z1 by z2 gives
= (cos(θ1 − θ2 ) + i sin(θ1 − θ2 )). Hence
has modulus
z2
r2
z2
r1
(divide the moduli) and argument (θ1 − θ2 ).(subtract the arguments)
r2
It follows by induction that if z = r(cos θ+i sin θ) then for n ∈ N, z n = rn (cos nθ+i sin nθ);
this is called De Moivre’s Theorem and in fact it is true for all n ∈ R.
1
1
1
1
= (cos θ − i sin θ) and n = n (cos nθ − i sin nθ).
z
r
z
r
De Moivre’s Theorem allows us to calculate many useful trigonometrical identities in a
straightforward manner
If z = r(cos θ + i sin θ), then
10
Example 6
Find cos 3θ and sin 3θ in terms of powers of cosines and sines respectively.
1.6
Sets of complex numbers
We can describe lines and curves in the complex plane by defining conditions on z: if
these conditions are inequalities we have a description of a region of the complex plane.
• The distance between two points in the complex plane, z and w is |w − z|, thus the
set {z : |z| = a} is a circle centre at the origin and radius a.
• The set {z : |z − b| = a} is a circle centre (b, 0) and radius a. The set {z :
|z − (b + ci)| = a} is a circle centre (b, c) and radius a.
• The set {z : |z − ai| = |z − bi|} is the set of points that are the same distance from
ai as they are from bi - they lie on the perpendicular bisector of ai and bi.
• arg(z − (a + bi)) = ψ is a line passing through the point (a, b) and making an angle
of ψ with the positive x-axis.
• The set {z : |z| < 3} is all of the points that lie inside a circle centre the origin and
radius 3 - excluding the boundary. (This is called an open set).
• The set {z : a ≤ |z| ≤ b} comprises points that lie in an annulus bounded by two
circles, centre the origin, of radii a and b, including the boundary of the circles
themselves (a closed set).
• The set {z : − π4 ≤ arg(z) ≤ π4 } is all the points that lie in the triangular sector
shown below. (The diagonal boundaries are included but the set is open.)
11
imaginary
π/4
−π
/4
real
• The set {z : |z − (3 + i)| < |z + 1|} is the set of points that are nearer to 3 + i than
they are to −1. This is the set of points on the right hand side of the perpendicular
divisor of the line joining 3 + i and −1. The mid point of this line is (1, 21 i) and the
gradient of the line is 14 . Thus the points lie on the right hand side of
y−
1
= −4(x − 1) −→ 2y + 8x = 9
2
Roots of polynomial equations
The Fundamental Theorem of Algebra states that every polynomial equation with
coefficients in C has all its solutions in C
Complex
roots of real polynomials occur in conjugate pairs. Suppose that P (z) = a0 +
∑n
r
r=1 ar z . Then it is straightforward to show that P (z) = P (z̄). Now, if α is a root of
P (z) then P (α) = 0 so P (α) = 0 and thus P (ᾱ) = 0. hence if α is a complex root of a
polynomial with real coefficients then so is ᾱ.
Example 7
Given that x = 1 +
roots.
√
3i is a root of x4 − 5 x3 + 12 x2 − 16 x + 8 = 0, find the other three
12
Exercises
1. Express in the form a + ib, where a, b ∈ R,
(a) (2 – 3i) – (4 – 5i),
(b) (3 + 4i)(2 – 3i),
(c) (5 – i)2 ,
(d)
6 − 2i
.
3 + 4i
2. For each of the following, find (i) its modulus, (ii) its argument in radians between
−π and π, in terms of π or as a decimal.
√
(a) 1 + i,
(b) 3 – 4i,
(c) –2 + 5i,
(d) − 3 – i,
(e) –7i
(f) 7.
3. Find all the solutions in C of the equations (a) 4x2 + 1 = 0, (b) x2 + 2x + 5 = 0.
4. Prove from the definitions that for complex numbers w and z,
(a) w + z = w + z,
(b) wz = w z.
5. Show that z ÷ z has modulus 1. Express arg(z ÷ z) in terms of arg (z).
6. Find the quadratic equation which has 2 + 3i as one of its roots.
7. Suppose (a + bi)2 = 5 + 12i. By expanding the left-hand side and equating the real
and imaginary parts, find the possible values of the real numbers a and b. Hence
write down the two square roots of 5 + 12i.
Deduce the value of tan φ, if tan 2φ = 12/5 and 0 < φ < π.
8. If z = cos θ + i sin θ, expand z 4 by the binomial theorem. Hence express cos 4θ and
sin 4θ in terms of powers of sin θ and cos θ.
9. Find cos 7θ in terms of powers of cos θ
10. *Find (i) tan 3θ and (ii) tan 5θ terms of powers of tan θ
1
= 2i sin nθ.
zn
Deduce that 16 sin5 θ = sin 5θ − 5 sin 3θ + 10 sin θ.
11. If z = cos θ + i sin θ, show that z n −
12. *Express (i) cos6 θ (ii) cos7 θ in terms of cosines of multiples of θ.
13. Given that z = 2 + 3i is a root of the equation z 3 − 6z 2 + 21z − 26 = 0, find the
other two roots.
14. Sketch the following curves in the Argand diagram and give the Cartesian form of
the equation.
(a) |z + 2 − i| = |z − 1 + 2i|
(b) Re(z) = Im(z)
(c) |z − 2 + 3i| = 4
z−2
π
(d) *arg
=
z+5
4
13
15. Prove de Moivre’s Theorem by induction for n ∈ N.
16. Sketch the following regions in the Argand diagram, labelling clearly the boundaries
that are included
(a) Im(z) ≥ (Re(z))2
(b) {z : − π2 < arg(z) < − π6 }
(c) {z : |z| > 5|z + 6|}
(d) *{z : 3 ≤ |z − 2 + 3i| ≤ 4}
17. Show, geometrically or otherwise that
||z1 | − |z2 || ≤ |z1 − z2 | ≤ |z1 + z2 |
for any two complex numbers z1 , z2 .
√
18. Show that x = 2 + 3i is a root of
x4 − 8x3 + 26x2 − 40x + 21
and find the others.
14
Chapter 2
Functions
2.1
Domain and range
Let X and Y be sets. A function f : X → Y is a rule which assigns to each x in the
domain X exactly one element f(x) in the codomain Y . We call x the argument of the
function f . Note that a function is single valued - in other words any vertical line cuts
its graph no more than once.
If Y ⊆ R, f is a real-valued function. If Y ⊆ C, f is a complex-valued function.
In this course, X and Y will be subsets of R unless otherwise stated.
We write f : x 7→ f (x), where x ∈ X, and read this as f maps x to its image f (x).
√
If no domain is specified, we take f to have its maximal domain, e.g. f : x 7→ 1 − x2
can only be defined for −1 ≤ x ≤ 1. The complete definition of a function requires
that the domain is specified, thus f (x) = sin x, x ∈ [0, π] is not the same function as
f (x) = sin x, x ∈ R.
The subset of Y given by {f (x) : x ∈ X}, or f (X), is called the range of f . Thus
the range of a function is the set of y-coordinates at all the points on the graph of the
function.
2.1.1
Finding the range of a function
If f is quadratic, we can find its range by finding its minimum or maximum point, either
using calculus or by completing the square. f : x 7→ x2 + 4x − 3 ≡ (x + 2)2 − 7 has
range f (x) ≥ −7, or equivalently [−7, ∞). The minimum point on the graph of f (x) is
at (−2, −7).
For more general functions with domain R, the range can often be found by writing
f (x) = y and finding a condition for this to have real roots for x.
15
Example 8
Find the range of f (x) =
2.2
3x + 2
,
x2 + 4
Types of functions
In this section we consider a real-valued function f with domain X ⊆ R.
2.2.1
Odd and even functions
If f (−x) = f (x) for all x ∈ X we say that f is an even function. The graph of an even
function is symmetric about the y-axis.
If f (−x) = −f (x) for all x ∈ X then f is an odd function. The graph of an odd
function is the same when rotated by 180◦ about the origin.
Example 9
If f (x) is even and g(x) is odd, then:
f (x) × g(x) =
We have the following rules for functions
• odd ± odd = odd
• even ± even = even
• odd ± even = neither odd nor even
• odd × odd = even
• even × even = even
• even × odd=odd.
16
2.2.2
Periodic functions
We describe a function as periodic if, for some k > 0, f (x + k) = f (x) for all x ∈ X and
k ∈ Z. The smallest such positive k is called the minimal period. The function is said to
exhibit translational symmetry.
2.2.3
One to One functions
A one-to-one or injective function is such that if f (a) = f (b) then a = b, i.e. no two
elements of X have the same image under f .
Example 10
f (x) = x3 and g(x) = x4
Any horizontal line cuts the graph of a one to one function no more than once,
2.2.4
Onto functions
An onto or surjective function is one where the range of f is the whole codomain of
f . Thus for every element in the codomain there exists an element in the domain which
maps to it.
Example 11 f : R → R where f (x) = 5x + 2 and g : R → R where g(x) = x2 + 2x
It is often simpler to show that f is not onto by finding a counter example.
Any horizontal line cuts the graph of an onto function at least once.
2.2.5
Bijective functions
A function is bijective if it is both injective and surjective. Every horizontal line cuts
the graph of a bijective function exactly once.
17
2.2.6
Polynomials
A polynomial is a function of x of the form
a0 + a1 x + a2 x2 + ... + an xn ,
where a0 , ...an are constants. n is the degree of the polynomial.
2.2.7
Monotonicity
We say that f (x) is monotone increasing in some interval [a, b] if f (x1 ) ≥ f (x2 ) ⇔ x1 ≥ x2
for all x ∈ [a, b]. If the relationship is f (x1 ) > f (x2 ) ⇔ x1 > x2 then f is strictly monotone
increasing. If f (x1 ) ≥ f (x2 ) ⇔ x1 ≤ x2 then we say that f is monotone decreasing while
if f (x1 ) > f (x2 ) ⇔ x1 < x2 , then f is strictly monotone decreasing.
An equivalent definition is that if f (x) is monotone increasing in some interval [a, b]
then f 0 (x) ≥ 0 for all x ∈ [a, b].
Theorem
If f : A → B is strictly monotone, then f is one to one.
2.3
Some important functions
The identity function
The identity function id is defined by
id(x) = x for all x.
2.3.1
The absolute value function
The absolute value or modulus abs(x) or |x| is defined by
{
x,
x≥0
|x| =
−x,
x<0
√
√
Note that x always means the positive square root of x, and so |x| = x2 .
2.3.2
The signum function
The signum function, sgn is defined by

 −1,
0,
sgn(x) =

1,
18
x<0
x=0
x>0
√
x
x2
Thus for x 6= 0, sgn(x) =
and sgn(x) =
.
|x|
x
2.3.3
Integer part
The integer part or floor of x, denoted by bxc or [x], is defined to be the largest integer
less than or equal to x. For example, b7c = 7, b−2.3c = −3 and bπc = 3.
2.3.4
Heaviside function
We define the Heaviside function as
H(x) =
{
1,
0,
x≥0
.
x<0
The Heaviside function is a ”switch” which turns a function on at a particular value thus
the graph of f (x) = H(x − π) cos x starts at (π, −1).
2.4
Composition of functions
For functions f and g, the composite function g ◦ f, or gf, is defined by
(g ◦ f)(x) ≡ g(f(x)). To find this, substitute f(x) in place of x as the argument of g(x),
and simplify.
If f has domain X and g has domain Y , the domain of g ◦ f is {x ∈ X : f(x) ∈ Y }.
Similarly (f ◦ g)(x) = f(g(x)). In general g ◦ f 6= g ◦ f. However, composition is associative,
i.e. (f ◦ g) ◦ h = f ◦ (g ◦ h). Note that f ◦ id = id ◦ f = f.
19
Example 12
√
f (x) = 1 + x. and g(x) = cos x
It is useful to be able to identify functions as compositions,
Example 13
√
sin( x + 3)
2.5
Inverse functions
If f is bijective, with domain X and codomain Y , then there exists an inverse function
f −1 with domain Y and range X. The graph of f −1 (x) is the reflection of the graph of
f (x) in the line y = x.
f −1 is defined by the property
f ◦ f −1 = f −1 ◦ f = id,
(
)
i.e. f f −1 (x) = f −1 (f (x)) = x.
Sometimes the inverse can be found by inspection. Otherwise, make x the subject of y =
f(x) and then swap x and y in the answer. It is normal to use x in defining the inverse
function, but y or any other symbol is not wrong if it is used consistently.
20
Example 14
Find the inverse of f : R → R defined by f (x) = 1 + 3x.
A function whose graph is symmetric about y = x is self-inverse, i.e. f = f −1 , such as
x
.
f (x) =
x+1
If f is not bijective, we may obtain an invertible function by restricting the domain.
Example 15
f (x) = x2 − 8x + 11 = (x − 4)2 − 5
Theorem
If f : A → B and f is invertible then if f is strictly decreasing (increasing) f −1 is strictly
decreasing (increasing).
21
2.6
Partial fractions
p(x)
, where p(x), q(x) are polynomials.
q(x)
Many rational functions can be expressed in partial fractions. The principles are:
A rational function of x has the form
(i) A polynomial of degree n in the denominator, which does not factorise, requires a
polynomial of degree n − 1 in its numerator, i.e.
ax2 + bx + c
p
qx + r
≡
+ 2
.
2
(x + d)(x + ex + f )
x + d x + ex + f
(ii) If there is a linear factor to the power n in the denominator, there may be partial
fractions with all powers of this linear factor up to the nth in their denominators, i.e.
ax2 + bx + c
p
q
rx + s
+ 2
≡
+
.
2
2
2
(x + d) (x + ex + f )
x + d (x + d)
x + ex + f
(iii) If the degree of numerator ≥ degree of denominator, there will be some non-fractional
terms in the answer. In this case, do a long division first:.
Example 16
2x4
x3 + 4x2 + 3x + 12
We usually find partial fractions over Q. However, we can also have partial fractions over
R or C, e.g.
Example 17
x2
1
−3
22
Example 18
1
x2 + 4
23
Exercises 2(a)
1. State the maximal domains of the functions
√
(a) f : x 7→ x2 − 9,
(b) g : x 7→
x2
3x + 1
.
− 2x − 3
2. Find the range of the function f : R → R defined by f (x) = x2 − 4x − 3. Why does
f not have an inverse? How can the domain and codomain of f be restricted so that
the resulting function does have an inverse?
√
√
(−1)2 .
3. State the values of sgn(–3), | − π|, b 2c,
(iv) H(x−2)x2
√
5. If f : (−∞, 0) → (1, ∞) by x 7→ 1 − 5x and g : [1, ∞) → [0, ∞) by x 7→ x − 1,
define (if they exist) the functions f−1 , g−1 , f ◦ g and g ◦ f.
4. Sketch graphs of (i) sgn(1−x),
(ii) x−bxc,
(iii) sgn(sin x)
6. For each of the following functions with maximal domain, and codomain R, state
whether it is even, odd, periodic, one-to-one, onto. Find and simplify the composite
functions f ◦ f, f ◦ g, g ◦ f, g ◦ h, and state their domains.
2x
(a) f : x 7→ x2 + 1, (b) g : x 7→
, (c) h : x 7→ tan 2x,
x−2
2x + 1
for x ∈ R, find the set of values of y for which f(x) = y has
x2 + 2
real roots for x. Hence state the range of f.
7. If f (x) =
8. Sketch a graph of
{
f (x) =
2x,
x2 ,
.
−1 ≤ x < 0
0≤x≤1
Define the inverse function.
9. Show that if f ◦ g is invertible, then (f ◦ g)−1 = g −1 ◦ f −1 .
10. Express each of the following in partial fractions:
4x − 3
2x
(b)
(a)
(x + 1)(x2 + x + 1)
(x − 5)2 (x + 1)
(c)
x3 + 1
x2 + 7x + 12
(d)
3x + 7
(x + 1)2 (x + 3)
11. Express each of the following in partial fractions:
x2 + 2
3x4 + 6x3 − 2x2 + 4
(a) 5
(b)
4x + 4x3 + x
x3 + 2x2
12. *The functions f and g are defined as follows
{
{
x2 + 4 x ≥ 1
3x
x≥0
f (x) =
g(x) =
x
x<1
−2x x < 0
Find f ◦ g and g ◦ f .
24
13. Assume that f is an odd function and g is an even function, both defined on the
real line. Is each of the following functions odd, even or neither?
f ◦ g,
g ◦ f,
f ◦ f,
g ◦ g.
14. Find all the real values of the constants A and B such that f (x) = Ax + B satisfies
(i) f ◦ f (x) = f (x) for all x, (ii) f ◦ f (x) = x for all x.
15. *For what values of a, b and c is the function f (x) =
x−a
self inverse?
bx − c
16. Show that if f (x) is an odd function, its derivative is an even function.
17. *Express the following in partial fractions over C
√
3x − 4
−7 3 − ix
x2 − 2
(i) 2
(ii)
(iii)
x + 16
x2 + 3
x3 − 1
25
2.7
Trigonometric functions
In order that we can extend the trigonometric functions which are first encountered defined
in a right angle triangle to the whole real line, let P (x, y) be a point on the circle with
centre at the origin and radius 1, and let θ be the angle measured anti-clockwise from the
positive x-axis to OP . Then we define:
sin θ = y, cos θ = x, tan θ =
sin θ
y
= .
cos θ
x
The sine and cosine functions have range [−1, 1], so a function of the form r sin(nx + α)
2π
has range [−r, r]. r is the amplitude and the period is
. The phase is α
n
[
]
If f (x) = sin x for x ∈ − π2 , π2 then f is one-to-one. With this restriction of the domain,
the inverse trigonometric function arcsin x or sin−1 x is defined for −1 ≤ x ≤ 1. Similarly, restricting the domain of cos x to [0, π], the inverse(function
is arccos x or cos−1 x,
)
for −1 ≤ x ≤ 1. Also, restricting the domain of tan x to − π2 , π2 , the inverse function is
arctan x or tan−1 x, for x ∈ R.
Note that cos−1 x =
2.8
π
2
− sin−1 x and tan−1 x =
π
2
− cot−1 x.
The exponential and logarithmic functions
The exponential function exp(x) can be defined as the sum of the convergent series:
exp(x) = 1 + x +
x2 x3
+
+ ...
2!
3!
exp(1) is an irrational (cannot be expressed as p/q where p, q ∈ Z), transcendental (is
not the root of a polynomial equation with rational coefficients) number 2.718 2818 2845
9045 2354. . ., which we call e. It can be shown that exp(x) is this number to the power
x, so exp(x) is denoted by ex .
ex can also be defined in this way:
(
lim
n→∞
1+
x )n
= ex .
n
The exponential function is not bijective: for instance, there is no x ∈ R such that
ex = −1, showing that it is not surjective. However if the codomain is restricted to the
positive real numbers then the exponential function becomes bijective; its inverse is the
natural logarithm function ln x. Note that ln x is defined only for x > 0. Thus if y = ex
then x = ln y.
Other exponential functions can be defined on R, e.g. ax = exp(x ln a).
26
2.9
Hyperbolic functions
Any function f , defined on a domain which is symmetrical about the origin, can be
expressed as the sum of an odd function and an even function as follows :
1
1
f (x) ≡ (f (x) − f (−x)) + (f (x) + f (−x)).
2
2
Taking f (x) = ex , the odd and even components are the hyperbolic functions
sinh x ≡
ex − e−x
2
We also define
cosech x ≡
and
cosh x ≡
ex + e−x
.
2
1
sinh x
, tanh x ≡
,
sinh x
cosh x
and
1
cosh x
1
, coth x ≡
≡
.
cosh x
sinh x
tanh x
The hyperbolic functions have many properties similar to those of the trigonometric functions, but they are not periodic.
sech x ≡
For example, cosh2 x − sinh2 x ≡ 1;
compare this with
cos2 x + sin2 x ≡ 1.
1 − tanh2 x ≡ sech2 x, coth2 x − 1 ≡ cosech2 x.
Similarly,
Note the important identity
cosh x + sinh x = ex .
2.9.1
Osborne’s rule
To convert a trigonometric identity into a hyperbolic one, replace cos by cosh and sin by
sinh but whenever sin2 occurs either explicitly or implicitly (e.g. in tan2 ), change the sign.
1. sin(A + B) ≡ sin A cos B + cos A sin B becomes
sinh(A + B) ≡ sinh A cosh B + cosh A sinh B;
2. cos(A + B) ≡ cos A cos B − sin A sin B becomes
cosh(A + B) ≡ cosh A cosh B + sinh A sinh B;
tanh A − tanh B
tan A − tan B
becomes tanh(A − B) ≡
1 + tan A tan B
1 − tanh A tanh B
(Note the implicit product of sines in tan A tan B).
3. tan(A − B) ≡
Example 19
Prove that
tanh 2x =
2 tanh x
.
1 + tanh2 x
27
To solve hyperbolic equations and establish identities we can use the definitions in terms
of exponentials, or any of the standard identities.
Example 20
Solve cosh x + 2 sinh x = 6
The following inverse hyperbolic functions are defined on the given domains:
√
sinh−1 x = ln(x + x2 + 1) x ∈ R
cosh−1 x = ln(x +
tanh−1 x =
√
x2 − 1) x ∈ [1, ∞)
1 1+x
ln
2 1−x
28
x ∈ (−1, 1).
Exercises 2(b)
1. Find the amplitude and the period and the phase where appropriate of
(a) 3 sin 3x
(b) 8 sin 3x cos 3x
(c) 2 cos(2x + 4)
(d) 6 cos x + 8 sin x
2. Starting from the definitions, prove that cosh2 x − sinh2 x ≡ 1.
3. Solve, in terms of natural logarithms, (a) 4 sinh2 x = cosh2 x,
4. Find the exact value of arcosh
(b) 7 sinh x = 24.
13
.
12
5. Find the coordinates of any points of intersection of the curves y = cosh 2x and
y = 3 − 2 cosh x.
6. Find an identity relating coth2 x and cosech2 x. Hence solve coth2 x = 2cosech x.
7. Prove the expressions for the inverse hyperbolic functions given in the preceding
section
(
)n
8. Prove that limn→∞ 1 + nx = ex .
9. Find an expression for sech x as a function of x.
10. Prove the following identities:
π
1
=
x>0
x
2
(b) cos−1 (−x) = π − cos−1 x |x| ≤ 1
x+y
(c) tan−1 x + tan−1 y = tan−1
xy 6= 1
1 − xy
(a) tan−1 x + tan−1
29
2.10
Functions of a complex variable
In this chapter so far we have been considering functions with a real argument which
give a real output. We now brief consider some of the properties of complex functions
f : W → C, where W ⊆ C.
2.10.1
The complex exponential function
Many power series are valid for complex numbers also. In particular, if z ∈ C then the
series
∞
∑
xn
z2 z3
=1+z+
+
+ ...
n
2!
3!
n=0
converges for all z. We call its limit exp(z) or ez .
ez has the following properties
• ez ew ≡ ez+w ,
e−z ≡
1
.
ez
• eiθ = cos θ + i sin θ. The unit circle in the complex plane can thus be described by
the set {eiθ : − π < θ ≤ π}.
• eiπ + 1 = 0. This is Euler’s famous identity.
• ez = ez+2nπi , n ∈ Z. The complex exponential function is periodic.
• |ez | = eRe(z)
• ez = ex (cos y − i sin y) = ez̄ (The complex conjugate of the exponential is the
exponential of the complex conjugate).
We can thus express z = x + yi in the form reiθ where r = |z| and θ = arg(z).
Example 21
Express z = 1 +
√
3i in exponential form.
Since eiz = cos z + i sin z and e−iz = cos z − i sin z we have the following Euler relations:
)
)
1 ( iz
1 ( iz
e + e−iz , sin z =
e − e−iz ,
cos z =
2
2i
which give the following relationships between trigonometric and hyperbolic functions:
cosh z = cos iz,
cosh iz = cos z,
sinh iz = i sin z,
i sinh z = sin iz
Note that these relationships, which can be shown to hold for z ∈ C of course hold for
θ ∈ R.
30
De Moivre’s theorem in exponential form
De Moivre’s Theorem can be stated as
( iθ )n
re
= rn einθ .
This is valid for any n ∈ R.
2.10.2
The cube roots of unity
The cube roots of unity are the three roots of z 3 = 1; if we factorise z 3 − 1 we obtain
(z − 1)(z√2 + z + 1) =
0. We solve this equation to find the three cubes roots of 1, these are
√
3
3
1
1
1, − 2 + 2 i, − 2 − 2 i. They are usually denoted by 1, ω, ω 2 . They have some interesting
properties,
1
1 + ω + ω 2 = 0,
ω̄ = ω 2 ,
ω 2 = ω,
ω = 2.
ω
0i
We can write 1 as cos 0 + i sin 0, or e , as well as cos 2π + i sin 2π = e2πi and cos 4π +
i sin 4π = e4πi . More generally we have
1 = e0i+2πki , k ∈ Z
√
0iπ
2iπ
4iπ
Then from de Moivre’s theorem 3 1 = e 3 , e 3 , e 3 . The last expression is of course
e−2πi/3 . Thus
2π
2π
+ i sin
= ω,
e2πi/3 = cos
3
3
while
2π
2π
e−2πi/3 = cos
− i sin
= ω2.
3
3
The cube roots of unity divide the unit circle into three equal parts, they form the apices
of an equilateral triangle.
imaginary
ω
Real
ω
2
The nth roots of unity
In an analogous manner, the nth roots of unity are the vertices of a regular n-sided polygon with one vertex at 1.
The seventh roots of 1 are 1, e2πi/7 , e4πi/7 , e6πi/7 , e−6πi/7 , e−4πi/7 , e−2πi/7 .
31
Example 22
Find the fourth root of z =
2.10.3
√
3 − i.
The logarithm of a complex number
We define the logarithm of a complex number in the obvious way, if z = ew then w = ln z.
To find what we mean by ln z suppose that
z = r(cos θ + i sin θ) = eu+iv = eu (cos v + i sin v).
Then r = eu and v = θ + 2kπ, k = 0, ±1, ±2 . . ., thus
Log(z) = u + iv = ln |z| + i(θ + 2kπ), k = 0, ±1, ±2 . . . .
The complex logarithm is many valued, which is why we use the symbol Log, the principal
value has θ ∈ (−π, π] and is thus ln z = ln r + iθ.
Example 23
Find ln(4 + 5i).
The complex logarithm has the same properties as the real logarithm.
2.10.4
Real and imaginary parts of complex valued functions
We frequently want to find the real and imaginary parts of complex functions, in the form
f (z) = u(x, y) + iv(x, y). Here are some examples,
32
Example 24
Find the real and imaginary parts of ez ,
2.10.5
ln z,
e−z ,
2
sin z,
tan z.
Solving equations in complex functions
Example 25
Solve sin z = 4
33
Exercises 2(c)
1. Find the real and imaginary parts of the following functions
1
1
z + (ii) z 2 − 2
z
z
2. Find the real and imaginary parts of the following functions:
1
(i) cos z (ii) e z (iii) z z (iv) arctan(z)
3. Solve the following equations:
z+1
(i)
= 2 + 3i
z−1
(ii) z 2 + 2z = 2 + 4i
(iii) cos z = 5
(iv) cosh z = −2
4. Prove that ω̄ = ω 2
5. Find the principal values of the logarithms of (i) −5 (ii) 2 + 7i
6. Find the real and imaginary parts of (1 + i)( 1 + i).
7. *Prove that ii is real.
8. *Prove that ln(z 2 ) = 2 ln(z)
34
Chapter 3
Limits, graphs and equations
3.1
Limits
The limit of f (x) as x tends to a, written lim f(x), is a number ` such that we can make
x→a
f (x) as close as we like to ` by taking x very close (but not equal) to a.
For a continuous function
lim f (x) = f (a).
x→a
Some functions have different limits as x → a from below (from the left) and from above
(from the right). In such a case, lim f(x) does not exist.
x→a
Example 26
Consider the function defined as follows
{
f (x) =
3.1.1
x,
x − 1,
x<1
x≥1
A note about infinity
Strictly speaking, tan π2 = ∞ is incorrect, since ∞ is not a number. A correct way to
write this would be
limπ tan x = ∞.
x→ 2
limx→∞ x1
1
=0
or ∞
= 0 means that as x gets larger and larger,
as ∞ is not a number.
35
1
x
tends to 0. Do NOT write
1
0
=∞
1
= ∞, but this is just another was to state that the limit does not
x→0+ x
We can write lim
exist.
3.1.2
Rules for limits
If f(x) and g(x) both have finite limits as x → a, then
lim cf(x) = c lim f(x) for any c ∈ C
x→a
x→a
lim [f(x) + g(x)] = lim f(x) + lim g(x)(the limit of the sum is the sum of the limits)
x→a
x→a
x→a
lim [f(x)g(x)] = lim f(x). lim g(x),(the limit of the product is the product of the limits)
x→a
x→a
x→a
lim f(x)
f(x)
x→a
lim
=
, provided lim g(x) 6= 0.(The limit of the quotient is the quotient of
x→a g(x)
x→a
lim g(x)
x→a
the limits)
Example 27
lim
x2 + x − 2
.
x2 + 5x + 6
lim
axn + (a0 + a1 x + a2 x2 . . . + an−1 xn−1 )
bxm + (b0 + b1 x + b2 x2 . . . + bm−1 xm−1 ))
x→−2
x→∞
3.1.3
Some important limits for trigonometric functions
Geometrically (as well as by other methods) it can be shown that for any acute angle x,
sin x < x < tan x. Dividing through by sin x,
1<
1
x
<
,
sin x
cos x
Now as x → 0, cos x → 1, so also
lim
x→0
sin x
= 1,
x
so cos x <
sin x
< 1.
x
sin x
→ 1. It can be deduced that:
x
1 − cos x
tan x
lim
= 0,
lim
= 1.
x→0
x→0
x
x
36
3.1.4
Limits for exponential and logarithmic functions
We have seen that the exponential function can be defined in terms of a limit as follows:
(
x )n
ex = lim 1 +
.
n→∞
n
Letting n = p1 , we also have ex = lim(1 + px)1/p .
p→0
We can easily show that the inverse function of (1 + px)1/p is
of ex can also be expressed as a limit:
( p
)
x −1
ln x = lim
.
p→0
p
xp −1
,
p
so the inverse function
Many of the properties of logarithms can be deduced directly from this limit.
3.1.5
The Squeeze Theorem
Example 28
)
1
Find lim x cos x +
.
x→0
x
√
3.2
(
Curve Sketching
Sketching graphs of functions can provide a great deal of information about their behaviour. This is the process that should be followed when sketching the graph of y = f (x).
1. Intersections with the axes
Put y = 0, x = 0 to find the intersections with the x axis (the roots of f (x)) and
the y axis f (0).
37
2. Vertical asymptotes
When the denominator of a rational function approaches zero then y → ±∞. At
such a value of x there is a vertical line which the graph approaches but never meets.
Example 29
y=
x−3
(x − 1)(x − 4)
3. Horizontal asymptotes
Find lim f(x) to see what the graph is like when |x| is very large.
x→±∞
Example 30
y = 3 + e−x
4. Oblique asymptotes If the numerator is of higher degree than the denominator,
there may be oblique asymptotes. To find these, divide through.
Example 31
y=
x2 + 2x + 3
x−1
A graph may sometimes cross an oblique asymptote, but never a vertical nor a horizontal one.
In some cases the asymptote may be a curve.
Example 32
y=
x3 − x2 − 8
.
x−1
38
5. (Obvious) symmetries
If the function is even then there is symmetry about the y-axis. If the function
is odd then there is 180◦ rotational symmetry about the origin. If the function is
periodic the function will show translational symmetry. If the function is self-inverse
then there is symmetry about y = x.
6. Stationary points
Turning points can be found by the usual calculus methods: solve f 0 (x) = 0. The
second derivative f 00 (x) is positive at a minimum, negative at a maximum and 0 at
a point of inflexion.
3.2.1
Inequalities
Sketching the graph can often help with the solution of inequalities. To find where f (x) >
0, sketch f (x) and see where the graph lies above the x-axis.
To solve f (x) > g(x), either sketch both graphs and see where f (x) is above g(x), or
sketch f (x) − g(x) and find the values of x where this is positive.
Example 33 Solve cos x > 2 sin x.
39
3.2.2
Related graphs
Related graphs can be obtained in various standard ways. For example, to obtain y =
|f(x)|, simply reflect in the x-axis those parts of the graph that lie below it.
1
1
To get y =
note that
has asymptotes where f (x) = 0, and vice-versa.
f(x)
f (x)
3.3
Standard transformations
There are a number of standard (linear) transformations:
• y = f(x) + a : translate y = f(x) by a units parallel to the y-axis (upwards).
• y = f(x + a) : translate y = f(x) by −a units parallel to the x-axis (a to the left).
• y = f(−x) : reflect y = f(x) in the y-axis.
• y = −f(x) : reflect y = f(x) in the x-axis.
• y = kf(x) : stretch y = f(x) by a factor of k parallel to the y-axis.
• y = f(kx) : ‘stretch’ y = f(x) by a factor of
1
k
parallel to the x-axis.
Implicit Equations
Up to now we have looked at functions defined explicitly in the form y = f (x). However,
sometimes functions are given implicitly in the form f (x, y) = g(x, y).
Example 34
3y 2 + 4x5 y − 2x2 − 4 = 0.
40
We can try and spot forms that suggest standard curves such as conic sections, but in
many cases, e.g cos(x + y) = xey−x , we cannot obtain an explicit form nor easily find the
graph of the function. In fact the graph of this function is as below:
Sometimes a closed curve, which does not represent a true function but does express a
relationship between two variables, is best described implicitly.
The circle with centre (a, b) and radius r has equation
(x − a)2 + (y − b)2 = r2 .
Example 35
x2 + y 2 + 4x − 8y − 5 = 0.
3.4
Parametric Equations
It is often convenient to describe a curve (not necessarily representing a well-defined
function) by two equations giving each of x and y in terms of a variable parameter, say
t, in the form x = f (t), y = g(t).
41
Example 36 The circle x2 + y 2 = a2 .
Other common parametrisations are
• The ellipse
x2 y 2
+ 2 = 1 has parametric form x = a cos θ, y = b sin θ,
a2
b
θ ∈ [0, 2π)
x2 y 2
− 2 = 1 has parametric form x = a cosh t, y = b sinh t or
a2
b
x = a sec θ, y = b tan θ)
• The hyperbola
• The straight line passing through x0 , y0 and x1 , y1 can be parametrised as
x = x0 + t(x1 − x0 ), y = y0 + t(y1 − y0 ),
t ∈ [0, 1].
Many other types of curve can be described parametrically and this is a particularly useful
technique when the Cartesian equations of the curve would be extremely complicated.
Sketching a graph defined by parametric equations
The simplest way to do this is to eliminate the parameter if this is possible
Example 37 Sketch the curve defined by x = t2 − 1, y = t + 1.
If the parameter cannot be eliminated, an alternative way to sketch the curve is to work
out x and y for a suitable range of values of the parameter.
42
Example 38 x = t − 3 sin t, y = 4 − 3 cos t
3.5
Polar coordinates
Plane polar coordinates are an alternative set of coordinates to the usual Cartesian or
rectangular coordinates in two dimensions. We define the coordinates of a point as the
distance r from the pole (origin) and the angle θ (in radians) made with the initial line,
taken to be the positive x-axis. Then, in general if the polar coordinates of a point are
(r, θ), the points (r, θ ± 2kπ, k ∈ Z) also represent the same point. We allow r < 0 by
considering negative r to be measured from the pole opposite to the direction of positive
r. Thus the points (2, 4π
) and (−2, π3 ) are the same.
3
3.5.1
Relationship with rectangular coordinates
We have x = r cos θ, y = r sin θ, alternatively r =
Example 39
The straight line
43
√
y
x2 + y 2 , θ = tan−1 .
x
Example 40
The curve with polar form r = 2a cos θ
3.5.2
Drawing polar curves
As with parametric curves, if polar curves cannot be transformed into rectangular coordinates it is often necessary to plot some points on polar graph paper, which can be easily
constructed as below.
θ = π/2
θ = 3π/4
θ = π /4
r=4
r=3
r=2
r=1
θ =π
θ = 5π
/4
θ = 3 π /2
θ =0
θ = 7π
/4
However, for simpler functions the sketching process can be accomplished as follows:
1. Plot a graph of θ against r as if they were rectangular coordinates. This will show
clearly how θ varies with r.
2. Find the values of θ for which r = 0 and indicate them with rays, these are the
directions in which the curve approaches the origin.
3. Find maxima and minima of r to show where the curve is furthest away from and
nearest to the origin.
Example 41 Sketch the polar graph of r = 3 cos 2θ
44
Some common polar curves
• The equation r = c is a circle radius c, centre the origin.
• θ = c is a straight line from the origin at c radians to the initial line.
• A circle with centre at polar coordinates (a, ψ) has equation r = 2a cos(θ − ψ). In
general the polar graph of equation
r = f (θ − ψ) is the graph of r = f (θ) rotated anticlockwise by ψ.
• r = 2(1 − cos θ) is known as a cardioid, r2 = cos 2θ is a lemniscate
• r = θ is an equiangular spiral, r = e−θ is an exponential spiral
45
Note that whereas the curves in the previous item are 2π periodic, these spirals are
not periodic.
46
Exercises 3
1. Find the following limits:
2x2 + 3x
12x + 6
(a) lim
,
(b) lim
,
x→0
x→∞ 3x − 4
x
x−3
.
x→3 x2 − 9
(c) lim
2. The diameter of the pupil of an animal’s eye is given by f(x) mm, where x is the
intensity of the light on the pupil,
80x−0.3 + 60
where f(x) =
,
8x−0.3 + 15
find (i) the diameter of the pupil when there is no light,
(ii) the limit to which the diameter tends as the amount of light becomes large.
3. Find the√following limits or explain why they do not exist
x−3
(x − π)2
|x − 2|
(i) lim
, (i) lim
, (i) lim
x→9 x − 9
x→π
x→0 x − 2
πx
|x − 2|
x4 − 16
(i) lim
, (i) lim 3
.
x→2 x − 2
x→2 x − 8
f (x)
f (x)
= −2, find lim f (x) and lim
2
x→0 x
x→0
x→0 x
4. If lim
5. if 2 − x2 ≤ f (x) ≤ 2 cos 2x for all x, find lim f (x)
x→0
( )
1
6. *Find lim x sin
x→0
x
7. Sketch the following curves, and state the range of each function.
x
3x + 2
2
3x
(a) y =
, (b) y =
, (c) y =
, (d) y =
,
2
1+x
3 − 2x
1−x
1 − x2
(x + 1)(x − 2)
x2 + x − 1
x2 + 1
(e) y =
,
(f) y =
,
(g) y = 2
.
(x − 1)(x + 2)
x−1
x +2
8. Sketch the following curves
ln x
1 − x2
(a) y =
(b) y =
1 + |x|
1 + x2
2
1+x
x ln x
(d) y =
(c) y =
x(x + 1)(x − 1)
x + ex
9. Solve
x−1
x2 − 9
≥
.
x2 + x − 12
x2 − x − 2
10. f(x) is defined to be the larger solution (for y) of 2y 2 − 4xy − 3x4 = 0.
Find an explicit formula for f(x).
11. The circle C has equation x2 + y 2 − 6x + 8y − 144 = 0. Find the centre and radius
of C. Show that the point A (8, 8) lies on C and find the coordinates of B such that
AB is a diameter of C.
47
12. The parametric equations of a curve C are x = 1 + sinh t, y = 5 − 4 cosh t. Sketch
C for −1 ≤ t ≤ 1. Show that C meets the x-axis at two points, and state their
coordinates.
13. Convert the following polar equations to equations in rectangular coordinates:
(a) r = 3 sec θ
(b) r = sin θ + cos θ
2
(c) r =
2 − cos θ
(d) r = sec θ tan θ
14. Convert the following equations into polar form:
(a) y 2 = x(a − x)
(b) x2 + y 2 = 2a2 xy
15. Sketch the following polar curves
(a) r = 3(1 + sin θ)
(b) r2 = 16 sin 2θ
(c) r = 4θ,
(d) *r =
1
θ
θ ≤ 0 and r = 4θ,
and r =
1
.
θ−2
θ≥0
Can you explain your results?
16. Show that the distance between the two points (r1 , θ1 ) and (r2 , θ2 ) is given by
√
d = r12 + r22 − 2r1 r2 cos(θ1 − θ2 ).
17. Sketch the graph of y = (x − 4)3/2 .
48
Chapter 4
Differentiation
The gradient of the graph of a function at a given point describes the rate at which the
function is changing at that point. For example, if the function describes the displacement
of an object, then the gradient of a tangent to the graph gives its velocity.
Geometrically, the gradient of a graph at P (x, f (x)) is found by drawing a chord from
P to a nearby point Q = (x + δx, f (x + δx)), where δx means a small increment in the x
direction.
The gradient of this chord is
f (x + δx) − f (x)
.
δx
As δx → 0, the gradient of the chord tends to the gradient of the tangent.
Definition
Let f be a function defined in some neighbourhood of a point (x, f (x)). If the limit
lim
δx→0
f (x + δx) − f (x)
δx
exists then f is said to be differentiable at (x, f (x)). The value of the limit is the derivative
or derived function or differential coefficient of f (x) at this point, which we can write as
d
f 0 (x) or
(f (x)).
dx
f is a differentiable function if it is differentiable (i.e. the above limit exists) at
every point where it is defined.
If f (x) = y, the derivative is denoted by
δy
dy
, which can be understood as lim
.
δx→0 δx
dx
When δy and δx are corresponding small increments in y and x, δy ≈
formula is used to find the effect on y of known small changes in x.
49
dy
dx
δx. This
An alternative version of the definition, giving the derivative where x = a, is
f 0 (a) = lim
x→a
f (x) − f (a)
.
x−a
These formulae are used to differentiate functions from first principles.
Example 42
Differentiate f (x) = x3 from first principles.
4.1
Derivatives of Hyperbolic Functions
From the definitions in terms of exponentials, it follows that
d
d
(sinh x) = cosh x and
(cosh x) = sinh x.
dx
dx
Using these results, the quotient and composite function rules give the following standard
derivatives of hyperbolic functions:
f (x)
f 0 (x)
sinh(ax + b)
a cosh(ax + b)
cosh(ax + b)
a sinh(ax + b)
tanh(ax + b)
a sech2 (ax + b)
coth(ax + b)
−a cosech2 (ax + b)
sech(ax + b)
−a sech(ax + b) tanh(ax + b)
cosech(ax + b)
−a cosech(ax + b) coth(ax + b)
50
4.2
Differentiating Inverse Functions
If y is a function of u which is a function of x, the rule for composite functions states that
dy
dy du
=
·
.
dx
du dx
Putting u = x and x = y gives
dy dx
dy
=
·
dy
dx dy
so
dy dx
·
= 1.
dx dy
If y = f(x) is an invertible function of x, and f −1 (x) is easier to differentiate than f (x),
this technique can be used: express x in terms of y, differentiate with respect to y to get
dx
dy
1
and then use
=
.
dy
dx
dx/dy
(First derivatives obey the laws of fractions, higher derivatives do not.)
Example 43
y = sin−1 (2x − 3)
This method gives the following standard results for the inverse trigonometric and hyperbolic functions.:
f(x)
f 0 (x)
arcsin x ≡ sin−1 x
1
,
1 − x2
1
,
−√
1 − x2
1
1 + x2
1
√
1 + x2
1
√
x2 − 1
1
,
1 − x2
√
arccos x ≡ cos−1 x
arctan x ≡ tan−1 x
arsinh x ≡ sinh−1 x
arcosh x ≡ cosh−1 x
artanh x ≡ tanh−1 x
51
x ∈ (−1, 1)
x ∈ (−1, 1)
x ∈ (1, ∞)
x ∈ (−1, 1)
4.3
Parametric Differentiation
If x = f (t) and y = g(t), then
dy dt
dy/dt
g 0 (t)
dy
=
·
=
= 0 ,
dx
dt dx
dx/dt
f (t)
provided that f 0 (t) 6= 0.
Example 44 Find y 0 if x = a sec x and y = b tan x
For second derivatives, use
dy
dy dt
d
=
·
=
dx
dt dx
dt
(
dy
dx
)
·
dt
f 0 (t)g 00 (t) − f 00 (t)g 0 (t)
=
dx
(f 0 (t))3
Example 45 Find y 0 if x = t3 − 3t, y = t2
.
4.4
Implicit Differentiation
An implicit equation f (x, y) = g(x, y) can be differentiated term-by-term with respect
dy
, which can be abbreviated
to x. Whenever y occurs, it must be differentiated to get
dx
as y 0 .
Example 46
y sin x = x3 − cos y
To obtain second derivatives we differentiate through the equation again; y 0 differentiates
to y 00 .
52
Example 47
xy + y 2 = 2x
4.5
Logarithmic Differentiation
Taking logarithms of both sides of an equation sometimes helps in differentiation.
1 0
1 dy
=
f (x).
If y > 0, y = f (x) becomes ln y = ln f (x), which differentiates to
y dx
f (x)
Example 48
f (x) = ax
Example 49
y=
√
4.6
x sin xex ,
Leibniz’s Rule
This is a method for finding the nth derivative of a product of two functions.
53
Let f and g be n-times differentiable functions. Let h be defined by h(x) = f (x) g(x).
Then h is also n times differentiable and it can be proved that
n ( )
∑
n (r)
(n)
h (x) =
f (x) g (n−r) (x),
r
r=0
where h(n) (x) means the nth derivative of h(x) with respect to x. We can also use D the
dn
differential operator to express this, where n f (x) = Dn (f (x)).
dx
Example 50
d3 (x3 e2x )
dx3
4.7
Derivatives of functions in polar coordinates
Consider a function r = f (θ) in plane polar coordinates. Then x = f (θ) cos θ and y =
f (θ) sin θ. Thus we have
dy
= f (θ) cos θ + f 0 (θ) sin θ
dθ
and
dx
= f 0 (θ) cos θ − f (θ) sin θ.
dθ
dy
Solving for the values of θ at which
= 0 will give the points at which the tangent to
dθ
dx
the curve is horizontal and at which
= 0 the points at which it is vertical. In general
dθ
we have
f (θ) cos θ + f 0 (θ) sin θ
dy
= 0
.
dx
f (θ) cos θ − f (θ) sin θ
54
Example 51 Find the points where the tangent lines to the cardioid r = 1 + cos θ are
horizontal and vertical.
4.8
Applications of differentials
You should be familiar with the application of differentiation to gradients, tangents, normals, turning points, maxima and minima and rates of change.
55
4.8.1
Radius of Curvature
Another feature of a graph that can be found by differentiation is the radius of curvature
of y = f (x), defined as
3
[
( )2 ]
2
dy
1+
dx
2 ρ=
.
d y
dx2 1
The curvature is defined to be κ = .
ρ
Example 52
Find the radius of curvature of y = cosh x at x = ln 2
4.8.2
l’Hôpital’s Rule
Suppose f and g are differentiable real-valued functions with f(a) = g(a) = 0. Then
(
)
f(x) − f(a)
f(x)
f(x) − f(a)
x−a
lim
= lim
= lim
.
x→a g(x)
x→a g(x) − g(a)
x→a
x−a
g(x) − g(a)
for x 6= a
f(x) − f(a)
f 0 (a)
x→a
x−a
=
= 0
g(x) − g(a)
g (a)
lim
x→a
x−a
lim
provided g0 (a) 6= 0.
1 − sin x
x→ 2
cos x
Example 53 limπ
If f0 (a) =g0 (a) = 0 we can repeat the process to get
f 00 (a)
f(x)
= 00 ,
x→a g(x)
g (a)
lim
provided g00 (a) 6= 0.
56
1 − cos x
x→0
x2
Example 54 lim
If lim f (x) = ∞ and lim g(x) = ∞ then l’Hôpital’s rule is expressed in the form:
x→a
x→a
f (x)
f 0 (a)
= 0
x→a g(x)
g (a)
lim
x
x→∞ ex
Example 55 lim
4.8.3
Related rates of change
If x and y are both functions of t then if z = f (x, y) we have
dz
df dx df dy
=
+
dt
dx dt
dy dt
Example 56
If the sides, and of a rectangle are 25 cms and 40 cms and are increasing at 2 cm per
second and decreasing at 3 cm per second respectively find the rate of change in the area,
57
4.8.4
Small changes in the dependant variable
We use the equation ∆y ≈
dy
∆x.
dx
Example 57
Find the change in the period of a simple pendulum relative to a change in its length
Exercises 4
1. Differentiate from first principles : (a) x5 ,
(b)
1
,
x
(c) sin 2x.
2. Differentiate with respect to x:
(a) (x2 + 3)5/2 , (b) sinh(cosh x), (c) arcsin
( x+3 )
2
(d) arctanh (1 − x2 ),(e) coth2 (ex ),(f) arcsech x (i.e. sech−1 x)
3. *Differentiate with respect to x:
√
(
)
cosec x
1 − 2x2
(2x + 3)3
−1
(b)
cot
(c)
(a) sin
1 + 2x2
x3 + 5
(4x2 − 1)8
Hint: use logarithmic differentiation.
4. Use logarithmic differentiation to find the derivative with respect to x of
x tan2 x
, where x > 1.
(a) (sin x)x ,
(b) 3cosh x ,
(c)
x3 − 1
√
1
1 + x(1 − x) 3
0
5. Use logarithmic differentiation to compute the value of y where y =
4
(1 + 5x) 5
6. Express
d2 x
dy
in terms of
.
2
dy
dx
7. Find the maximum and minimum values of sin−1 (x2 − 1) for −1 ≤ x ≤ 1.
8. A curve has parametric equations x = a(1 − cos 2t), y = a(2t + sin 2t), where a is a
d2 y
π
non-zero real constant and 0 ≤ t ≤ 2 . Find
when t = π4 .
2
dx
58
9. A circle has equation x2 + y 2 − 2x + 6y − 15 = 0. Find the gradient of the tangent
to this circle at the point where x = 5 and y < 0,
(a) by implicit differentiation,
(b) by finding the centre and radius and using coordinate geometry.
10. The implicit equation of a curve is 2y 2 − 3xy + x = 6. Find equations of the tangent
and the normal to the curve at the point where x = 1 and y > 0.
d2 y
dy
.
Also find
in
terms
of
x,
y
and
dx2
dx
11. Given that f (x) = 13 sinh x(2 + cosh2 x), show that f 00 (x) = cosh3 x.
Find, to the nearest integer, the radius of curvature of the curve y = f(x) at
the point where x = ln 2.
12. Use Leibniz’s Rule to find
(a) the third derivative of x4 cos 2x,
(b) an expression for the nth derivative of x3 sinh x.
13. Use l’Hôpital’s rule to find the following limits:
x2 − a2
sin πx
(a) limx→a 3
,
(b) limx→2 2
,
3
x −a
x −4
ln x
limx→1 x
.
e −e
(c)
limx→0
arctan x
,
ln(x + 1)
(d)
14. The equation of curve S is x3 + y 3 − 6y − 9x = 0
(a) Find the co-ordinates of the points where S intersects the x axis.
(b) Find the equation of the normals to S when x = 3.
(c) Find the x co-ordinates of the turning points of S.
(d) Find the values of y at which the curve is vertical.
(e) What is the equation of the oblique asymptote of S?
(f) Sketch the curve.
15. Prove Leibnitz’s rule. hint: use induction.
16. *Find the points at which the tangents to the following polar curves are horizontal
or vertical. Sketch the curves
(i) r = sin 2θ (ii) r = eθ (iii) r = sin θ cos2 θ
17. Determine the equation (in rectangular coordinates) of the tangent line to
r = 3 + 8 sin θ at θ = π6 .
59
Chapter 5
Series
5.1
Power series
A series of the form
a0 + a1 (x − c) + a2 (x − c) + . . . =
2
∞
∑
ar (x − c)r
r=0
is called a power series in x − c or a power series about the point x = c. c is the
centre of convergence of the series. For the values of x for which the series converges
the sum defines a function of x. The series converges provided that |x − c| < R, i.e.
c − R ≤ x ≤ c + R where R is the radius of convergence,given by
an+1 1
.
R=
and L = lim n→∞
L
an Note that if L = 0, R = ∞ and if L = ∞, R = 0. The series obviously converges for x = c
if nowhere else.
Example 58 Find the centre, radius and interval of convergence of the series
∞
∑
(2nx + 5)n
n=0
60
3n (n2 + 1)
.
5.1.1
Operations on power series
A convergent power series in x (i.e. with c = 0) or x − c is simply a function of x. Thus
if f (x) is represented by the convergent power series P (x) then f (g(x)) is represented by
P (g(x)). The interval of convergence of the original series was |x| < R, of the new series
it is the interval such that |g(x)| < R.
Example 59
Find the binomial expansion for
1
1 + x2
61
The Cauchy product
We can add and subtract convergent series to obtain another series with radius of convergence at least as large as the smaller of the radii of convergence of the original series.
We can multiply two convergent series together, to obtain the Cauchy product thus
(∞
)( ∞ )
∞
∑
∑
∑
an
bn =
cn
n=0
where
cn =
n
∑
n=0
n=0
aj bn−j = a0 bn + a1 bn−1 + . . . + an b0 .
j=0
Example 60
Find the series expansion for
5.1.2
1
by direct multiplication.
(1 − x2 )2
Differentiating power series
We treat a convergent power series as an infinite polynomial in x and differentiate it term
by term to obtain another series with same radius of convergence.
Example 61
find the series expansion for
1
by differentiation.
(1 − x2 )2
62
5.1.3
Integrating power series
We can integrate a convergent power series term by term, making sure to evaluate the
constant of integration, and obtain another series with same radius of convergence.
Example 62
Find a power series expansion for tan−1 x about x = 0.
5.2
Taylor series
It is straightforward to show that, provided f [n] (c) exists for all n ∈ N, we can represent
f (x) as a power series in x − c:
f (x) = f (c) + f 0 (c)(x − c) +
f 00 (c)(x − c)2 f 000 (c)(x − c)3
+
+ ....
2!
3!
This is called the Taylor series of f about c. If c = 0 it is known as a Maclaurin series.
The series will converge provided that |x| < R, where R is the radius of convergence:
(n + 1)f [n] (c) .
R = lim n→∞
f [n+1] (c) Remark
We can also write the Taylor series in the form
f (x + h) = f (x) + hf 0 (x) + h2
f 000 (x)
f 00 (x)
+ h3
...
2!
3!
This form has important applications in approximating functions and in the numerical
solution of differential equations.
5.2.1
Some common Maclaurin series
Here are some standard Maclaurin series. They are convergent for all real x unless otherwise stated. Note that in trigonometric series, x must be in radians.
63
∑ xr
x2 x2 x2
• e =1+x+
+
+
+ ... =
2!
2!
2!
r!
r=0
∞
x
∑
x3 x5
x2r+1
+
− ... =
(−1)r
• sin x = x −
3!
5!
(2r + 1)!
r=0
∞
2r
∑
x2 x4
r x
• cos x = 1 −
+
− ... =
(−1)
2!
4!
(2r)!
r=0
∞
∑ x2r+1
x3 x5
• sinh x = x +
+
+ ... =
3!
5!
(2r + 1)!
r=0
∞
∑ x2r
x2 x4
+
+ ... =
2!
4!
(2r)!
r=0
∞
• cosh x = 1 +
∑
x 2 x3
xr
• ln(1 + x) = x −
+
− ... =
(−1)r−1
2
3
r
r=1
∞
∑ xr
x2 x 3
• ln(1 − x) = −x −
−
− ... =
2
3
r
r=1
(−1 < x ≤ 1)
∞
(−1 ≤ x < 1)
Note that ln x itself has no Maclaurin series, as it is not defined at x = 0.
Maclaurin series as its first derivative is not defined at x = 0.
5.2.2
√
3
x has no
Obtaining Taylor and Maclaurin series
While directly evaluating the derivatives of the function whose Taylor expansion is required will always give the series, it is often much simpler to manipulate other, known
series, by multiplication, division, differentiation and integration of those series as well as
simply by substituting for the argument. The following series are convergent for all real
x unless otherwise stated.
Example 63
x2
Find the Maclaurin series for e− 2 .
64
Example 64 Find the Maclaurin series for tan x.
Example 65 Find the Maclaurin series for sin2 x.
Example 66 Find the Taylor series for ln x about x = 3.
5.2.3
The remainder term
Truncating the Taylor series for f (x) around c at the term in (x − c)n will leave an error
term. This term can be written as the Lagrange remainder, En :
En =
f [n+1] (ξ)(x − c)n+1
(n + 1)!
where ξ ∈ [c, x].
65
Example 67
Find the error in calculating e−0.5 if the Maclaurin series is truncated at the term in x4
66
Exercises 5
1. Find the sums of the following series
∞
∞
∑
∑
5
1
(i)
(ii)
103n
(2 + π)2n
n=0
n=5
(iii)
∞
∑
3 + 2n
n=0
3n+2
.
2. Find the centre, radius of convergence and interval of convergence of the following
power series:
(i)
∞
∑
3n(2x − 1)n
(ii)
n=0
∞
∑
n3 (2x − 3)n .
n=0
3. Use multiplication of series to find a power series representation of
1
for |x| < 1.
(1 − x)3
4. Find the radius of convergence and the sum of the following infinite series
• 1 − 4x + 16x − 64x . . . =
2
3
• 3 + 4x + 5x2 + 6x3 . . . =
• 2 + 4x2 + 6x4 + 8x6 =
∞
∑
(−1)n (4x)n
n=0
∞
∑
(n + 3)xn
n=0
∞
∑
2(n + 1)x2n .
n=0
5. Find Maclaurin series for the following functions and state the radius of convergence.
(i) cos 3x3
(v) sec x
(ii) sin x cos x
(iii) ln(2 + x2 )
(iv) cosh x − cos x
6. *Find Maclaurin series for the following functions and state the radius of convergence
(i) esin x
(ii)
1
1 + x + x2
(iii)
1
.
1 + tan x
7. By direct multiplication of the series for ex and ey show that ex+y = ex ey .
8. Find the Taylor series for the following functions and state the interval of convergence:
1
(i) ex about x = 4
(ii) sin x about x = π4
(iii) 2 about x = −2.
x
9. Estimate the error if the Maclaurin series for the following functions are terminated
as indicated:
(i) sin 0.2, terminated with the term in x5
(ii) ln(cosh 2), terminated with the term in x4 .
67
10. *Find the following
∞
∑
sin nθ
(i)
,
n!
n=0
∞
∑
cos nθ
(ii) *
n
n=1
68
Chapter 6
Integration
6.1
The anti-derivative
There are two (equivalent) ways of defining integration. The first is as the inverse of
differentiation. Given a function f , we∫look for a function F such that F 0 (x) = f (x). If
this can be found, we denote F (x) by
f (x)dx. This is an indefinite integral.
Example 68
If F (x) = tan−1 ex find f (x)
F (x) is sometimes called an anti-derivative or primitive of f (x).
Clearly if F (x) is an anti-derivative of f (x) then so is F (x) + c for any real constant
c.
6.2
The definite integral
The second approach uses area. If f is defined on the interval (α, β) and the area enclosed
by its graph and the x-axis between x = α and∫x = β is finite, this area is called the
β
definite integral of f (x) from α to β, denoted by
f (x) dx and we then say that f(x) is
α
integrable on (α, β).
69
This yields the definition of a definite integral as
∫
β
f (x) dx = lim
N →∞
α
with δx =
6.3
N
∑
f (α + iδx)δx
i=0
β−α
. Note that area below the x-axis is regarded as negative.
N
The Fundamental Theorem of Calculus
The relation between definite and indefinite integration is given by this important theorem.
Let f : (α, β) → R be an integrable function. (as defined in the previous section)
Then there exists a function F : (α, β) → R such that F 0 = f, and
∫
β
f (x) dx = F (β) − F (α).
α
It follows that
d
dx
∫
x
f (t) dt =
α
d
(F (x) − F (α)) = f (x).
dx
Differentiating an integral
Example 69
d
Find
dx
∫
x
e−t dt
a
More generally
d
dx
∫
f (x)
h(t) dt = h(f (x))f 0 (x) − h(g(x))g 0 (x)
g(x)
70
Example 70
d
Find
dx
6.4
∫
x4
ln t dt
x2
Properties of Definite Integrals
∫
∫
β
•
β
f (x) dx, for a ∈ R
af (x) dx = a
α
∫
α
∫
β
•
(f (x) ± g(x)) dx =
α
∫
∫
f (x) dx
β
α
∫
• If f is odd,
f (x) dx
α
γ
f (α − x)dx
0
∫
α
f (x) dx = 0.
−α
∫
• If f is even,
β
f (x) dx +
α
f (x) dx =
0
∫
γ
f (x) dx =
α
∫
α
∫
β
• If α ≤ γ ≤ β then
•
g(x) dx
α
β
f (x) dx = −
∫
β
f (x) dx ±
α
α
•
∫
β
∫
α
f (x) dx = 2
−α
α
f (x) dx
0
∫
∫
α
• If f is periodic with period k and N ∈ Z then
f (x) dx =
0
∫
∫
β
• If f (x) ≤ g(x) and α < β then
f (x) dx
Nk
β
f (x) dx ≤
α
N k+α
g(x) dx
α
∫
x
The natural logarithm function can be defined as ln x =
1
1
dt.
t
Sometimes relationships between functions can be deduced from different forms of the
same indefinite integral.
71
Example 71
Show arcsin x + arccos x =
6.5
π
2
by integration.
Integration techniques
An essential part of integration techniques is to be able to process integrands so that they
resemble functions whose integrals are known. Often we cannot integrate directly, but
have to transform an integral into a recognisable standard form. In this respect, integration is much less routine than differentiation which is a very straightforward process.
Indeed many quite simple functions cannot be integrated to give closed form functions.
They are however typically integrable - i.e. the definite integral can be evaluated numerically.
You should always verify your answers by differentiation to obtain the integrand.
6.5.1
Standard Integrals
Here a and b are real constants. In each case a constant of integration, c should be added
for indefinite integrals.
72
f(x)
sinh(ax + b)
cosh(ax + b)
ax (a > 0)
1
a 2 − x2
1
2
a + x2
1
√
2
a + x2
1
√
,
2
x − a2
√
6.5.2
∫
f(x) dx
1
cosh(ax + b)
a
1
sinh(ax + b)
a
ax
ln a
x
arcsin
x ∈ (−a, a)
a
x
1
tan−1
a
a
x
sinh−1
a
x
cosh−1
x ∈ (a, ∞), a > 0
a
Integration by Substitution
In the substitution method we choose a suitable variable u = g(x), so du = g 0 (x) dx, and
obtain the integral of a function of u with respect to u.
Thus
∫
∫
(
)
du
f (x)dx = f g −1 u 0 −1 .
g (g u)
Having integrated, the answer must then be expressed in terms of x.
The limits on a definite integral can be converted into values of the new variable.
∫
π
cos3 x dx
Example 72
π/2
Rational functions containing square roots of quadratic functions can often be converted
(e.g. by completing the square) to one of the following types:
√
• If a2 − x2 occurs, try substituting
x = a sin u, so dx = a cos u du.
Simplify using a2 (1 − sin2 u) ≡ a2 cos2 u.
73
√
• If x2 − a2 occurs, try substituting
x = a cosh u, so dx = a sinh u du , since cosh2 u − 1 ≡ sinh2 u,
or x = a sec u, so dx = a sec u tan u du, since sec2 u − 1 ≡ tan2 u.
√
• If x2 + a2 occurs, try substituting
x = a sinh u, so dx = a cosh u du, since sinh2 u + 1 ≡ cosh2 u,
or x = a tan u, so dx = a sec2 u du, since tan2 u + 1 ≡ sec2 u.
(Sometimes one alternative will work better than the other.)
∫
1
1
1
√
dx, first put x = , so dx = − 2 du, etc.
For integrals of the form
u
u
x ax2 + bx + c
∫
∫
√
√
1
x
√
√
Also be aware of:
dx = 2 x + a + c,
dx = x2 + a + c.
x+a
x2 + a
1
The substitution t = tan x
2
x
2t
If t = tan , then tan x =
, from which we get
2
1 − t2
2t
1 − t2
2 dt
sin x =
,
cos
x
=
, and dx =
.
2
2
1+t
1+t
1 + t2
∫
1
This is useful for some integrals of the form
dx, which are transa sin x + b cos x + c
formed into polynomials.
Example 73
∫
Evaluate
dx
cos x + 2 sin x + 1
Forms to recognise
These forms are important to be able to recognise, doing so greatly simplifies the computation of the integral.
74
∫
•
∫
•
∫
•
∫
•
∫
•
6.5.3
f 0 (x)
= ln |f (x)| + c
f (x)
f 0 (x)f (x) =
1
(f (x))2 + c
2
f 0 (x)ef (x) = ef (x) + c
g 0 (x)f 0 (g(x)) = f (g(x)) + c
f 0 (x)
−1
2 = tan (f (x)) + c
1 + (f (x))
Integration by Parts
This method is obtained from the product rule for differentiation:
d
dv
du
dv
d
du
(uv) = u + v , hence u
=
(uv) − v .
dx
dx
dx
dx
dx
dx
Integration of this formula gives
∫
∫
∫
∫
dv
du
u
dx = uv − v
dx, or, equivalently
u dv = uv − v du.
dx
dx
Integration by parts is normally used for a product in which one factor, such as a power
of x, becomes simpler or disappears when differentiated. More than one application of
the method may be needed.
Sometimes ‘multiplication by 1’ is used to express a single function as a product, i.e.
f(x) = f(x) × 1. when we know the differential but not the integral of a function.
Example 74
∫
Find
sin−1 x dx
75
Miscellaneous examples
∫ 3
x +2
Example 75
dx
x2 + 2
∫
Example 76 Find
1
dx.
x(x2 + 1)2
∫
Example 77 Find
√
−3x2
1
dx.
+ 12x − 8
76
∫
Example 78
x2
∫
Example 79
Example 80
Example 81
1
dx
x2 + 1
sin2 x
dx
1 + cos x
∫ √
∫
√
4 − x2 dx
1
dx
cos4 x
77
6.6
Reduction Formulae
There are many cases in which it is useful to reduce an integral
involving a power of some
∫
n
function to one ∫involving a lower power, e.g. if In = sin x dx we can express In in
terms of In−2 = sinn−2 x dx . Such a reduction formula is commonly, but not always,
found using integration by parts.
∫
Example 82 Find
sin5 x dx
78
6.7
Expressions that cannot be integrated to closed
form functions
There are many expressions which cannot be integrated to simple closed form functions.
The include such simple looking integrals as
∫
∫
∫
∫
ln x
sin x
−x2
e
dx,
dx,
dx, ln(sin x)) dx.
x
x
Numerical techniques can be used to evaluate definite integrals of these expressions. We
may be able to use a Taylor or Maclaurin series to do this.
∫
0.2
cos(x2 ) to ten decimal places.
Example 83 Evaluate
0
79
6.8
6.8.1
Integration and polar curves
Length of a polar curve
It is straightforward to show that the length of a continuous polar curve r = f (θ) as θ
increases from a to b (provided no segment is traced more than once) is given by
∫
L=
b
√
(f (θ))2 + (f 0 (θ))2 dθ.
a
Example 84 Find the total length L of the cardioid r = 1 + cos θ is traced out as θ
moves from zero to 2π
80
6.8.2
Areas bounded by polar curves
For a continuous curve r = f (θ), the infinitesimal area δA bounded by the rays θ = a and
θ = b is a segment of a circle radius r with angle δθ = b − a, thus
∫ b 2
∫ b
δθ 2
r
(f (θ))2
δA =
πr −→ A =
dθ =
dθ.
2π
2
a 2
a
Example 85 Find the area within the first quadrant of the cardioid r = 1 − cos θ.
Sketching the curve will frequently show how symmetry may be used to simplify the
calculation of areas.
Example 86
(A method of finding the intersection of polar curves)
Find the points of intersection of the two cardioids r = 1 − cos θ and r = 1 + cos θ.
6.8.3
Volumes bounded by polar curves
For completeness we note that the volume of the solid obtained by rotation of the area of
the continuous curve r = f (θ) lying between θ = a and θ = b around the initial line is
∫
2π b
(f (θ))3 sin θdθ.
V =
3 a
In practice such volumes are more easily obtained by double integration (covered in the
Spring module)
81
6.9
6.9.1
Applications of integration
Mean value
The mean value of f (x) over the interval [α, β] is equal to
∫ β
1
f (x) dx.
β−α α
6.9.2
Length of arc
The length of the arc joining the points on y =f (x) at which x = α and x = β is
√
( )2
∫ β
dy
dx.
s=
1+
dx
α
6.9.3
Surface Area of a solid of rotation
If this arc is rotated once about the x-axis, the curved surface area is
√
( )2
∫ β
dy
A = 2π
y 1+
dx.
dx
α
6.10
Improper Integrals
It can happen that the definite integral of a function is finite even though x or y tends to
∞ or −∞ at one of the limits of integration or at some value between them.
A limit of integration is infinite
∫
∫
∞
We define
β
f (x) dx to be lim
α
β→∞
f (x) dx, if this limit exists and is finite. If the limit
α
does not exist or is infinite then the integral diverges.
∫ ∞
dx
Example 87
3
1 x
82
The function does not have a finite value at one of the limits of integration
∫
∫
β
If f (x) does not have a finite value at x = α we define
f (x) dx to be lim
α
k→α
β
f (x) dx,
k
if this limit exists. If the limit does not exist or is infinite then the integral diverges.
∫
Example 88
0
1
√
1
1−x
The function is not defined at some value within the interval of integration
∫
4
dx
has a discontinuity at x = 2 and must
2/3
1 (x − 2)
therefore be evaluated as the sum of two improper integrals.
Example 89 Note that the integral
83
Exercises 6
1. Find the integrals
∫
(a)
sinh(3x − 4) dx
1
5
√
dx
4 − x2
∫ 4
(b)
0
∫
(d)
∫
(x2
1
dx
+ 2x + 2)
(e)
3
4x2
∫
(c)
3x +
sin x
dx
cos x + 1
1
dx
− 20x + 25
2. Using suitable substitutions, or otherwise, find the integrals
∫
∫
x−2
1
√
√
dx
(b)
(a)
dx
x2 − 1
(x − 1)(x − 3)
∫
∫
x+1
x
√
√
dx
(d)
(c)
dx
2
2
1−x
x −x+1
∫ √ 2
∫ √
x −9
(e)
dx
(f)
2x − x2 dx
x
3. Find the integrals:
∫
cos x
(a)
dx
sin x + cos x
∫
(b)
1
dx
2 − sin x
∫
1
(c)
−1
√
(1 − x2 ) dx
4. Using partial fractions or otherwise, find the integrals
∫
∫
x2
x+1
(a)
dx
(b)
dx
(x − 1)(x − 3)
x(x + 5)2
∫ 3
∫
x + 4x2 − x + 3
x3
dx
(d)
dx .
(c)
(x + 1)(x2 − 4)
(x − 2)(x2 + 1)
∫ π
2
1
π
dx =
.
5. Prove that, for a, b > 0,
2
2
2
2
2ab
0 a cos x + b sin x
1
over the interval 0 ≤ x ≤ 6. By reference
x2 + 5x + 6
to a sketch, explain why the mean value you have found is reasonable.
6. Find the mean value of y =
7. A curve is given by the parametric equations x = 2 sinh3 t, y = 3 cosh2 t, for
0 ≤ t ≤ ln 3. Find the total length of the curve.
8. The arc of the curve y = cos x between x = 0 and x = π2 is rotated once about the
x-axis. Find the area of the curved surface formed.
∫ 1
9. Given that In =
xn cosh x dx, where n is a positive integer, prove that for n ≥ 2,
0
In = sinh 1 − n cosh 1 + n(n − 1)In−2 .
Hence find, in terms of hyperbolic functions, the value of I4 .
∫
10. Prove that if In = secn xdx, then
84
• I0 (c) = x + c
(
)
cos x/2 + sin x/2
• I1 (x) = ln
cos x/2 − sin x/2
• (n − 1)In (x) = tan x secn−2 x + (n − 2)In−2 (x),
∫
Using this, calculate sec5 x
x ≥ 2.
11. Evaluate the following, if they exist.
∫ ∞
∫ 2
∫ −2
1
−1
−1/3
(a)
dx
dx
(b)
x
dx
(c)
2
x3
1
0
−∞ x + 4
∫ π/2
∫ 4
1
√
(d)
tan x dx
(e)
dx.
x2 − 4x + 3
0
3
12. *The Gamma function Γ(x) is defined as the improper integral
∫ ∞
tx−1 e−t dt x ∈ R, x > 0
Γ(x) =
0
(a)
(b)
(c)
(d)
(e)
Find Γ(1).
Prove that Γ(x + 1) = xΓ(x)
Using the results in the previous two parts find Γ(2), Γ(3), Γ(4)
√
Given that Γ( 12 ) = π, find Γ( 32 ), Γ( 52 )
What is Γ(n) for n ∈ N?
∫ +∞ dx
13. Sketch the region whose area is 0 1+x
2 and show that this area can also be
∫ 1 √ 1−y
expressed as 0
dy
y
14. *Integrate the following:
∫ √
∫
∫
∫ √
1
(i)
x2 − 4xdx (ii)
dx (iii) sin(ln(x))dx (iv) e x dx
6
x +x
∫
∫
∫
∫
√
dx
dx
2
−1
(v)
ln(1
+
x
)dx
(vi)
tan
xdx
(vii)
1 + x + x2 + x 3
x ln x
15. Evaluate
∫ e the following:
∫ 4
(i)
| ln(x)|dx (ii)
√
1/e
0
dx
√
x+ 1+x
16. Find the lengths of the following polar curves:
(i) r = e4θ , θ ∈ [0, 2] (ii) *r = 1θ , θ ∈ [ π4 , π2 ]
In (ii) what can you say about the arc length of the portion of the curve that lies
inside the circle r = 1?
17. Find the areas enclosed by the following polar curves:
(i) the cardioid r = 2 + 2 cos θ
(ii) the inner loop of the limaçon r = 1 + 2 cos θ Hint: r ≤ 0 over the interval of
integration.
√
(iii) the area enclosed by the intersection of the circles r = 4 cos θ and r = 4 3 sin θ
(iv) the lemniscate r2 = sin 2θ. Hint:use the symmetry of the curve.
85
18. Find the volume of the solid produced when the cardioid r = 1 + cos θ is rotated
about the initial line.
19. The polar curve r = f (θ) can be parametrised as x = f (θ) cos(θ), y = f (θ) sin(θ).
Derive the formula for arc length of a polar curve.
86
Chapter 7
First order ordinary differential
equations
7.1
Types of first order o.d.e.s
First order ordinary differential equations (o.d.e.s) contain only first derivatives, y 0 . The
degree of the differential equation is the highest power of the derivative that appears.
Thus (y 0 )2 = x2 y is a first order o.d.e. of second degree. Where we are given the values of
the variables at some point we refer to the o.d.e. as an initial value problem (IVP). We
call the differential equations ordinary because there are no partial derivatives.
7.1.1
Solving a first order o.d.e
Solving the first order o.d.e. y 0 = 2y gives y(x) = Ce2x and thus gives rise to a family
of solutions depending on the parameter C. In this case y is defined for all x ∈ R and
y ∈ (0, ∞). Thus every point (x, y) lies on a solution curve, moreover it lies on only one
solution curve. To find a specific, unique solution we need only specify a particular value
of (x, y) such as y(x0 ) = y0 . Thus y 0 = 2y, y(0) = 4 has the solution y(x) = 4e2x .
Always check that your solution is correct by verifying by differentiation and rearrangement that it satisfies the original differential equation
Many o.d.e.s will also have y = constant as a solution (e.g. y 0 = 2y has y ≡ 0). This is
known as the trivial solution, you should therefore always check to see if there is a trivial
solution to the o.d.e. you are solving.
7.2
Variables separable equations
These have the general form
dy
f (x)
=
.
dx
g(y)
87
dy
If we rearrange the equation to give g(y)
= f (x) and then integrate both sides with
dx
respect to x we have
∫
∫
g(y)dy = f (x)dx.
Then we make y the subject of the equation. In principle this type of o.d.e. can always
be solved, provided that f and g can be integrated to give closed form functions.
Example 90
dy
= 5t4 y 2
dt
y(0) = 1
Example 91
dy
= cos x tan y.
dx
7.3
Linear equations - the integrating factor method
The equation must be in the following specific form (it is said to be linear in y and y 0 );
dy
+ yP (x) = Q(x).
dx
We first transform the equation into one where the variables are separable and then apply
the method of the previous section. The steps are as follows
88
1. Make sure the the o.d.e. is in the specific form above (by rearranging or substitution
if necessary).
2. Work out the integrating factor e
tion at this stage).
R
P (x)dx
(Note that there is no constant of integra-
3. Multiply both sides of the equation by the integrating factor,
e
R
P (x)dx dy
dx
+ ye
R
the left hand side can be written as
differentiating products).
P (x)dx
P (x) = Q(x)e
R
P (x)dx
d ( R P (x)dx )
ye
(check this by using the rule for
dx
4. Now the equation is
R
d R P (x)dx
ye
= Q(x)e P (x)dx .
dx
5. Integrate both sides with respect to x
∫
R
R
P (x)dx
ye
= Q(x)e P (x)dx dx + c.
6. Make y the subject of the equation
∫
y(x) =
,
R
Q(x)e P (x)dx dx + c
R
.
e P (x)dx
Example 92
ty 0 = y + t3
89
Example 93
Solve the IVP y 0 + 2xy = 4x y(0) = 3.
7.4
Homogeneous equations
These are equations where each of the polynomial expressions have the same order (e.g.
(y)
dy
2 3
4
x y and xy are both of order 5). They can be rearranged into the form
.
=H
dx
x
y
dy
du
With the substitution u = or y = ux we have
= u+x
by the chain rule. Thus
x
dx
dx
we can rewrite the equation as
du
u+x
= H(u)
dx
which we can rearrange to give us a variables separable equation
∫
∫
du
dx
du
= H(u) − u ⇒
=
x
dx
H(u) − u
x
which we now integrate and complete the solution by substituting for u.
Example 94 2xyy 0 − y 2 + x2 = 0
90
7.5
Bernoulli equations
These take the specific form:
dy
+ yR(x) = y n S(x).
dx
The right hand side contains y n so that the integrating factor method will not work
without further transformation of the equation. We now make the substitution u = y 1−n
with
dy
dy
y n du
du
= (1 − n)y −n
⇒
=
.
dx
dx
dx
1 − n dx
We can now transform the original equation in x and y into an equation in x and u as
follows;
y n du
+ R(x)y = S(x)y n
1 − n dx
1 du
+ R(x)y 1−n = S(x)
1 − n dx
⇒
1 du
+ R(x)u = S(x)
1 − n dx
du
+ u(1 − n)R(x) = (1 − n)S(x).
dx
The o.d.e. is now in the correct form to be solved by the integrating factor method.
Example 95 x
dx
+ y = x2 y 2 ln x
dy
91
7.6
Equations that can be transformed to one of the
types above
y
Just as the substitution u =
can be used to transform a homogeneous equation into
x
one that can, in principle, be solved, other substitutions may also be used to transform
what appear to be intractable equations, as the following miscellaneous examples show;
Example 96
y0 =
y t−1
+
t
2y
92
Example 97
y 0 = y − 4t + y 2 − 8yt + 16t2 + 4
Example 98
dy
y−x+1
=
dx
y+x+5
93
Example 99
dy
1
=
dx
x+y
Exercise
As an exercise, check the four solutions above.
7.7
Second order equations which can be solved as
first order equations
Provided that if y 00 and y 0 appear in the equation, y does not, we can make the substitution
y 0 = p(x) then y 00 = p0 and solve in the normal manner.
Example 100
d2 y dy
+
= sin x
dx2 dx
94
Example 101
d2 y
y 2 =
dx
(
95
dy
dx
)2
7.8
First order o.d.e.s with degree greater than 1
First order o.d.e.s of second and third degree cannot in general be solved by any of the
methods in this chapter. However, we can use the substitution y 0 = p. and attempt to
solve for p in some cases:
Example 102
(
dy
dx
)2
+ (x + y)
We can sometimes solve for x
dy
Example 103 y = 3x + 6y 2
dx
(
dy
dx
)2
96
dy
+ xy = 0.
dx
On the other hand we can solve for y.
(
2
Example 104 16x + 2y
dy
dx
)2
(
−x
dy
dx
)3
=0
97
7.9
Using Taylor series to obtain a series solution
Consider the differential equation (IVP) where we cannot obtain a solution by any of the
known methods, but we do know the value of the solution at some value of x.
dy
= f (x, y),
dx
y(x0 ) = y0 .
The Taylor series for y about x = x0 is
(x − x0 )2 d2 y (x − x0 )3 d3 y dy +
...
y = y0 + (x − x0 ) +
dx x0
2!
dx2 x0
3!
dx3 x0
and we use this to establish a series solution.
Example 105
dy
= x2 + y 2
dx
98
7.10
Modeling with first order odes
We consider some examples of the use of first order o.d.e.s to build simple but useful
models of the real world.
7.10.1
Population modeling
dN
If the population is N (t) the simplest model is
= ρN , where ρ is the difference
dt
between the per capita birth and death rates. If the population at some time t0 is N0 the
model gives us
N (t) = N0 eρ(t−t0 ) .
This very simple model is a useful way to study the early stage of the growth of cancer
cells, if r < 0 the cells die out. If r > 0 the cancer grows exponentially.
For animal and insect populations a better model recognises that resources are finite
and the o.d.e is now
(
)
N
dN
= rN 1 −
.
dt
K
Here r is the birth rate and K is the so-called carrying constant for the environment. We
dN
= 0 (i.e. the system is at equilibrium) when N = 0 (no population) and
see that
dt
N = K, the maximum population that the environment can support.
99
We can solve this o.d.e (left as an exercise) to give the logistic model:
N0 K
N (t) =
.
(K − N0 )e−r(t−t0 ) + N0
Note that, for N0 6= 0, lim Nt→∞ = K and that in theory it takes an infinite time to reach
this equilibrium.
7.10.2
Carbon dating
dM
Radioactive decay is modelled (quite precisely) by
= −kM where M (t) is the amount
dt
of radioactive material and k is a known decay constant specific to the isotope concerned.
If the amount present at t = t0 is M0 the model is
M (t) = M0 e−k(t−t0 ) .
Knowing the amount of radioactivity (e.g. 14 C) that would have been present in the wood
in a tree when it was cut down, the present level of radioactivity and the rate of decay of
14
C it is a simple matter to calculate the age of a wooden object.
7.10.3
Forensics - time of death
dθ
Newton’s law of cooling for a body at a temperature of θ(t) gives
= k(θ − θc ), where
dt
θc is the ambient temperature and k is a constant specific to the material of which the
body is made. This gives
θ(t) = θc + θ0 e−kt
where θ0 is the temperature at t0 . Knowing the temperature of the body, that of the
environment and the normal body temperature of 37.4C allows an estimation of the time
of death.
7.10.4
Finance - the continuous compounding of money
If the interest rate paid on the amount of money in an account A(t) is r per year, compounded over n periods each year then the amount of money in the account after m years
is
(
r )mn
Am = A0 1 + e n
.
As n → ∞ we know that this is Am = erm . From this we have that
dA
= −rA.
dt
We can say that an amount A0 invested at r continuously compounded will be worth
A(t) = A0 ert at time t. Conversely an amount of money A received at time t is worth
Ae−rt at today’s value (the present value). The present value P of an annuity of A0 per
year paid for n years will be the solution of
)
A0 (
dP
= A0 e−rt −→ P (t) =
1 − e−rt .
dt
r
100
7.10.5
Free fall with air resistance
Th downward force on a falling body is mg, where g is gravity and m is mass. The air
resistance is proportional to the velocity v so it is kv. Then from Newton, the downward
force on the body is
dv
dv
k
m = mg − kv −→
+ v = g.
dt
dt m
Then
)
g(
v(t) =
1 − e−kt ,
k
so that terminal velocity occurs when v = kg . We find the distance fallen by solving
)
)
dx
g(
gt
g (
=
1 − e−kt −→ x(t) =
− 2 1 − e−kt .
dt
k
k
k
101
Exercises 7
Solve the following differential equations. Verify that your solutions do indeed solve the
given o.d.e.s
e−2x
y2
(b) xy 0 = (x − 1)y y(1) = 1
√
(c) 1 + t2 y 0 = ty 3 y(0) = 2
1. (a) y 0 =
(d) y 0 = y 2 − 1 y(0) = 2
(e) sin xy 0 = 2y cos x
(f) yy 0 = 2(xy + x)
(g) yex+y y 0 = 1
2(y 2 + y − 2)
x2 + 4x + 3
(i) y 00 + (y 0 )2 + 1 = 0 (Hint; substitute y 0 = u)
(h) y 0 =
(j) xy 00 = y 0
(k) 2xy 0 + y = 0,
y(4) = 1
0
(l) (1 − x )y + 4xy = 0,
2
y(0) = 2
2. (a) y 0 + y = 5ex
(b) xy 0 + y = x4 − x y(1) = 2
2y
=1
(c) y 0 +
x+1
(d) ty 0 + 2y = et
(e) xy 0 = 2y + x2
(f) y 0 + 2xy + x = e−x
2
(g) y 0 + y tan x = sec x
(h) x2 y 0 + 2xy − x + 1 = 0
(i) (1 − x2 )y 0 + xy = 2x
y
= x2 − x
(j) y 0 +
1−x
(k) xy 0 + (1 + x)y = e−x
(l) y + y 0 = ex ,
y(0) = 2
(m) (1 + x2 )y 0 = 1 + xy,
0
y(1) = 0
(n) xy + x − 3y = 0 with (a) y(1) = −1 and (b) y(−1) = 1
2
(o) y 2 + (3xy − 4y 3 )y 0 = 0 (Hint; consider x as the dependent variable.)
3. (a) x2 yy 0 = x3 − y 3 y(1) = 1
√
(b) xy 0 = y + x2 + y 2 , y(4) = 3
102
2x − y
x − 2y
x+y
(d) y 0 =
x−y
x−y+5
(e) y 0 =
(Hint; substitute x = p + A, y = q + B where A and B are
x+y−1
suitable constants to transform this into a homogeneous equation.)
2x + 2y − 1
(f) y 0 =
3x + y − 2
(c) y 0 =
4. Use the substitutions given to solve the following equations for y
(a) y 0 = (y − t)2 − (y − t) − 1 let u = y − t
t2
yt e 2
(b) y =
+
let u = y 2
2
2y
y
y
y
(c) y 0 =
− + t2 (1 + t) let u =
1+t
t
1+t
0
5. Find the series solutions in ascending powers of x up to and including the term in
x3 of
dy
= y cos x, y(0) = 1
dx
dy
(b)
= exy , y(0) = 1
dx
(a)
In the first case compare your solution with the result of solving the o.d.e.
by direct integration and explain your conclusions.
6. Solve the following o.d.e.s. with the substitution u = x + y :
1
(x + y)2
(b) y 0 = sin(x + y)
(a) y 0 =
(c) y 0 = (1 − x − y) cos x − 1
√
(d) y 0 = x + y
7. Solve the following o.d.e.s:
8. Solve the following o.d.e.s with the substitution p =
(
)2
dy
dy
(a) xy
+ (x2 + xy + y 2 ) + xy = 0
dx
dx
( )2
dy
dy
(b)
− 2 cosh x + 1 = 0
dx
dx
( )3
dy
(c) y = x +
dx
103
dy
.
dx
dy
(d)
dx
(
dy
y+
dx
)
= x(x + y)
(
)
dy
dy
dy
(e) y = x + a
1−
dx
dx
dx
104
Chapter 8
Introduction to Fourier Series
8.1
Definition of a Fourier series
While Taylor and Maclaurin series are very useful polynomial approximations of functions,
they do not approximate periodic functions effectively. For this we use Fourier series.
Consider f (x) where f is 2π periodic, i.e. for all x we have f (x) = f (x + 2π). We
can approximate f (x) by
a0 ∑
f (x) =
+
(an cos nx + bn sin nx) ,
2
n=1
∞
where
1
a0 =
π
∫
π
f (x)dx,
−π
∫
1
an =
π
π
1
cos nxf (x)dx bn =
π
−π
∫
π
sin nxf (x)dx.
−π
(The derivation of the Fourier series will be covered in the lectures). If the function is 2L
periodic then with a scale change we have
a0 ∑ (
nπx
nπx )
f (x) =
+
an cos
+ bn sin
,
2
L
L
n=1
∞
where
1
a0 =
L
∫
L
f (x)dx,
−L
1
an =
L
∫
L
nπx
1
cos
f (x)dx bn =
L
L
−L
∫
L
sin
−L
nπx
f (x)dx.
L
a0 , an and bn are known as Fourier coefficients.
Recap - Products of odd and even functions
Suppose that f (x) is even and g(x) is odd. Then f (x) = f (−x) and g(x) = −g(−x) so
that f (x)g(x) = f (−x) (−g(−x)) = −f (−x)g(−x). Thus f (x)g(x) is odd.
105
If both f (x) and g(x) are even, then f (x) = f (−x) and g(x) = g(−x) so that f (x)g(x) =
f (−x)g(−x) and so f (x)g(x) is even.
Lastly, if both f (x) and g(x) are odd then f (x) = −f (−x) and g(x) = −g(−x) so
that f (x)g(x) = (−f (−x)) (−g(−x)) = f (−x)g(−x) and so f (x)g(x) is even.
The rule is odd×odd=even, even×even=even and odd×even=odd. This rule helps considerably in reducing the work involved in computing Fourier series.
8.2
Integrating odd and even functions
∫
If f (x) is even then
∫
If g(x) is odd then
∫
L
−L
L
−L
L
f (x)dx.
f (x)dx = 2
0
g(x)dx = 0 so a0 = 0 in this case.
Now, if f (x) is even, then cos
nπx
nπx
f (x) is even while sin
f (x) is odd.
L
L
Thus, for an even function:
∫
∫
nπx
nπx
1 L
2 L
cos
cos
f (x)dx =
f (x)dx,
L −L
L
L 0
L
and
1
L
∫
L
sin
−L
nπx
f (x)dx = 0.
L
The Fourier series of an even function contains only the cosine terms.
nπx
nπx
g(x) is odd while sin
g(x) is even, hence
L
L
∫
∫
∫
1 L
2 L
1 L
nπx
nπx
nπx
f (x)dx =
f (x)dx, and
g(x)dx = 0.
sin
sin
cos
L −L
L
L 0
L
L −L
L
If g(x) is odd then cos
The Fourier series of an odd function contains only the sine terms.
In summary;
• if f is even then
nπx
a0 ∑
+
an cos
.
f (x) =
2
L
n=1
∞
• if f is odd then
f (x) =
∞
∑
n=1
106
bn sin
nπx
.
L
Example 106
Compute the Fourier series for
{
f (x) =
0 : x ∈ ((4j + 1), (4j + 3)]
1 : x ∈ (4j − 1, 4j + 1]
for all j ∈ Z.
107
Example 107
Find the Fourier series for the sawtooth wave function given by
[
]
g(x) = x − 2jL for each interval (2j − 1)L, (2j + 1)L , for all j ∈ Z.
108
Example 108
Find the Fourier series for
f (x) = x2 + x,
−π < x < π
109
f (x) = f (x + 2π
8.3
Complex Fourier series
We have derived the identities e±inx ≡ cos nx ± i sin nx and
cos nx ≡
)
1 ( inx
e + e−inx
2
sin nx ≡
)
1 ( inx
e − e−inx .
2i
We can therefore express the general term of the Fourier series for a 2π periodic function
f (x) as
) bn ( inx
)
an ( inx
an cos nx + bn sin nx =
e + e−inx +
e − e−inx .
2
2i
and we can write the right hand side as
cn einx + kn e−inx
1
1
where cn = (an − ibn ) and kn = (an + ibn ).
2
2
Hence we can calculate
∫ π
∫ π
1
1
cn =
(f (x) cos nx − if (x) sin nx) dx =
f (x)e−inx dx,
2π −π
2π −π
∫ π
∫ π
1
1
(f (x) cos nx + if (x) sin nx) dx =
f (x)einx dx.
kn =
2π −π
2π −π
Since k−n = cn we can therefore express the Fourier series as
f (x) =
∞
∑
n=−∞
inx
cn e
,
1
where cn =
2π
∫
π
f (x)e−inx dx.
−π
In some circumstances this form of the Fourier series is more convenient to work with
than the sine and cosine series, as the following example shows.
110
Example 109
We require the Fourier series of the function f (x) = ex−2jπx ,
x ∈ [(2j − 1)π, (2j + 1)π).
111
j ∈ Z for each interval
8.4
Fourier series for functions given on one interval
only
Many functions are only physically interesting for a specific interval, thus the vibration
of a violin string is only given over its length. The idea is to extend the function so
that it becomes periodic. The extensions can be either odd (in which case we get the
sine terms) or even (in which case we get the cosine terms) and are known as half-range
extensions.This is illustrated in the following graph for the function f (x) = 4 sin x, x ∈
(0, π).
112
Example 110
Find the Fourier series for the odd and even half range extensions of the function
{
2x
0 < x ≤ L2
f (x) = 2L L
<L
L
2
113
8.5
Convergence of the Fourier series
We state the sufficient conditions for a function to have a convergent Fourier expansion,
they are known as Dirichlet’s conditions.
If f (x) is a bounded periodic function that in any period has a finite number of isolated maxima and minima and a finite number of finite discontinuities (i.e. it is bounded)
then the Fourier expansion of f (x) converges to f (x) at all points where f (x) is continuous and to the average of the right and left hand limits of f (x) at points where f (x) is
discontinuous.
Broadly speaking, the rate of convergence where the function has jump discontinuities
will be more rapid the “smoother” the function and in particular the rate of convergence
is slowest closest to a point of discontinuity. (This is known as the Gibbs phenomenon).
114
Exercises 8
1. Find the Fourier series of the following 2π periodic functions;
(a)
f (x) = x,
x ∈ (−π, π]
f (x) = x2 ,
x ∈ (−π, π]
(b)
(c)
{
f (x) =
(d)
{
f (x) =
−1 : x ∈ (−π, 0]
1 : x ∈ (0, π]
(
]
x : x ∈ (− π2 , π2]
1 : x ∈ π2 , 3π
2
(e) f (x) = ex , x ∈ (−π, π] (Use the real form of the Fourier series and compare
the difficulty with that using the complex form.)
2. Find the Fourier series of the following 2L periodic functions;
(a)
f (x) = |x|,
x ∈ (−2, 2],
3
(b)
f (x) =
πx
,
2
(c)
{
f (x) =
(d)
L=2
x ∈ (−1, 1],
1
2
1
2
L=1
(
]
+ x : x ∈ (− 21 ,]0
− x : x ∈ 0, 21
f (x) = π sin πx,
x ∈ (0, 1),
L=
L=
1
2
1
2
3. Find the complex Fourier series for the following 2π periodic functions;
(a)
f (x) = |x|,
(b)
{
f (x) =
x = (−π, π]
−1 : x ∈ (−π, 0]
1 : x ∈ (0, π]
4. f (x) = πx − x2 , x ∈ [0, π]. Find the Fourier series for both the odd and the even
half range extensions
{
π2
x ∈ (−π, 0)
5. f (x) =
and f (x + 2π) = f (x). Prove that the
2
(x − π) x ∈ (o, π)
Fourier series expansion is
]
∞ [
∞
(−1)n
4 ∑ sin(2n − 1)x
2 2 ∑ 2
cos nx +
π sin nx +
.
f (x) = π +
3
n2
n
π n=1 (2n − 1)3
n=1
115
Use this result to show that
∞
∑
1
π2
=
n2
6
n=1
and
116
∞
∑
(−1)n+1
π2
=
n2
12
n=1
SAMPLE CLASS TEST 1
You should show sufficient working to demonstrate the method you have used.
You have 45 minutes.
1. Find the four complex fourth roots of 2 + 2i in surd form and plot them in the
Argand diagram.
2. Sketch the following regions in the complex plane, indicating clearly which boundaries are included and which excluded.
(i) 4 < |z| ≤ 2
(ii) |z − 1| < |z − i|.
3. Find the range of the function f : R → R defined by f (x) = x2 + 6x − 4. By
restricting the range appropriately find the inverse function. Prove that your answer
is correct.
4. Starting from the definitions prove
(i) sinh−1 x = ln(x +
√
1 + x2 )
5. (i) Find the real and imaginary parts of e−z
(ii) sinh 2x = 2 sinh x cosh x.
2
(ii) Solve the equation cosh z = −2.
6. Find the following limits
1−x
√
lim
x→1 1 −
x
√
x4 + 4x
(ii) lim √
x→∞ 3 2x6 + 1
117
SAMPLE CLASS TEST 2
You should show sufficient working to demonstrate the method you have used.
You have 45 minutes.
1. This is the graph of y 3 x3 − x4 − y 4 + yx = 0.
(a) Explain the shape of the graph.
dy
3 y 3 x2 − 4 x3 + y
=
.
dx
4 y 3 − x − 3 y 2 x3
(c) Find the gradients of the two tangents to the curve when x = 12 , expressing
your answer in exact (i.e. surd) form.
(b) Show that
(d) Find the area enclosed by one of the loops. Hint: Find the equations of the
upper and lower curves in a loop.
[25]
2. Find the following limit: lim
√
x→∞
x
√
x
Hint: use l’Hôpital’s rule on the logarithm.[10]
3. Show that the Maclaurin series for sec x is
sec x = 1 +
∫
1 2
5 4
x +
x ....
2
24
0.1 sec(x2 )dx correct to 4 decimal places.
Hence find an approximation to
0
4. Evaluate the following integrals:
∫
(a)
dx(1 + x2 ) tan−1 x.
∫ √
(b)
x2 + 4 x − 1dx
118
[15]
∫
(c)
∫
(d)
3 x2 + 4 x − 1
dx.
x3 − 7 x + 6
sin ln x dx.
[25]
5. Solve the following ordinary differential equations:
dy
= y.
dx
1 dy
(b) 2
+ xy = x.
x dx
1 dy
(c)
+ x2 = x2 y 2 .
xy dx
(a) sin x
[15]
119
Calculus Exam Autumn 2009
Answer all five questions, they carry equal marks.
1. (a) Find the Maclaurin series for cos2 2x up to and including the term in x6 .
Hint: Start with an appropriate trigonometric formula.
(7)
√
(b) Explain why no Maclaurin series exists for (i) ln x and (ii) x.
(5)
∫ 1
(c) By finding the Maclaurin series for sin(x2 ), evaluate
sin(x2 )dx accurate to
0
3 decimal places.
(8)
2. (a) Use de Moivre’s Theorem to find trigonometric identities for cos 3x in terms of
cos x and sin 3x in terms of sin x.
(14)
∫ π
3 (
)
(b) Hence, or otherwise evaluate
cos3 x + sin3 x dx exactly.
(6)
π
6
3. Evaluate the following integrals
∫ π
4
sin x
(a)
dx
0 sin x + cos x
∫ 1
(b)
ln xdx
Hint: substitute u = tan x
(14)
Hint: Use l’Hopital’s rule
(6)
0
4. Solve the following ordinary differential equations
(a)
(1 + x2 )
(b)
e−x
(c)
(d)
dy
= x + xy 2
dx
dy
+ 2y = 1
dx
dy
+ y = y 3 ex
dx
d2 y
dy
+
3
= 4e−3x
dx2
dx
(5)
Hint: integrating factor method (5)
1
y2
dy
Hint: substitute u =
dx
Hint: substitute u =
(5)
(5)
5. Find the Fourier series for the following period 4 function:
{
x : x ∈ [0, 2)
f (x) =
1 : x ∈ [−2, 0)
(20)
120